ࡱ>  Y bjbj[[ 89 \9 \{H <<      8T %\ "t<+4p:":::4hY[[[[[[LH[ .fg@[<<::+ܺ <.: :Y Y j:pS<;E0"ORB( Hڌ č[[" > :*Mc lc  TOC \o "1-3" \h \z \u  HYPERLINK \l "_Toc236050159" Chng 1. Thut ton  PAGEREF _Toc236050159 \h 4  HYPERLINK \l "_Toc236050160" 1. V d  PAGEREF _Toc236050160 \h 4  HYPERLINK \l "_Toc236050161" 2. Khi nim  PAGEREF _Toc236050161 \h 4  HYPERLINK \l "_Toc236050162" 3. Cc c trng ca thut ton  PAGEREF _Toc236050162 \h 5  HYPERLINK \l "_Toc236050163" 4. Phng php biu din  PAGEREF _Toc236050163 \h 5  HYPERLINK \l "_Toc236050164" Bi tp luyn  PAGEREF _Toc236050164 \h 8  HYPERLINK \l "_Toc236050165" Chng 2. Gii thiu ngn ng lp trnh C++  PAGEREF _Toc236050165 \h 10  HYPERLINK \l "_Toc236050166" 1. Lch s hnh thnh  PAGEREF _Toc236050166 \h 10  HYPERLINK \l "_Toc236050167" 2. c im  PAGEREF _Toc236050167 \h 11  HYPERLINK \l "_Toc236050168" 3. Cu trc ca mt chng trnh C++  PAGEREF _Toc236050168 \h 12  HYPERLINK \l "_Toc236050169" 4. Mt s v d mu  PAGEREF _Toc236050169 \h 13  HYPERLINK \l "_Toc236050170" 5. Ci t chng trnh  PAGEREF _Toc236050170 \h 15  HYPERLINK \l "_Toc236050171" 7. Son tho chng trnh  PAGEREF _Toc236050171 \h 19  HYPERLINK \l "_Toc236050172" 8. Thot khi chng trnh  PAGEREF _Toc236050172 \h 20  HYPERLINK \l "_Toc236050173" Chng 3. Cc thnh phn v cc kiu d liu c bn  PAGEREF _Toc236050173 \h 21  HYPERLINK \l "_Toc236050174" 1. Cc thnh phn  PAGEREF _Toc236050174 \h 21  HYPERLINK \l "_Toc236050175" 2. Cc kiu d liu cn bn  PAGEREF _Toc236050175 \h 21  HYPERLINK \l "_Toc236050176" 3. Hng  PAGEREF _Toc236050176 \h 22  HYPERLINK \l "_Toc236050177" 3.1. Khi nim  PAGEREF _Toc236050177 \h 22  HYPERLINK \l "_Toc236050178" 3.2. nh ngh)a hng  PAGEREF _Toc236050178 \h 22  HYPERLINK \l "_Toc236050179" 4. Cc php ton ca C++  PAGEREF _Toc236050179 \h 23  HYPERLINK \l "_Toc236050180" 5. Xut nhp d liu  PAGEREF _Toc236050180 \h 24  HYPERLINK \l "_Toc236050181" 5.1. Xut d liu (cout)  PAGEREF _Toc236050181 \h 24  HYPERLINK \l "_Toc236050182" 5.2. Nhp d liu (cin)  PAGEREF _Toc236050182 \h 26  HYPERLINK \l "_Toc236050183" Chng 4. Cc cu trc iu khin  PAGEREF _Toc236050183 \h 27  HYPERLINK \l "_Toc236050184" 1. Lnh n v lnh phc  PAGEREF _Toc236050184 \h 27  HYPERLINK \l "_Toc236050185" 1.1. Lnh n  PAGEREF _Toc236050185 \h 27  HYPERLINK \l "_Toc236050186" 1.2. Lnh phc hay khi lnh  PAGEREF _Toc236050186 \h 27  HYPERLINK \l "_Toc236050187" 2. Cu trc iu kin if..else  PAGEREF _Toc236050187 \h 27  HYPERLINK \l "_Toc236050188" 2.1. Dng khuyt  PAGEREF _Toc236050188 \h 27  HYPERLINK \l "_Toc236050189" 2.2. Dng y   PAGEREF _Toc236050189 \h 28  HYPERLINK \l "_Toc236050190" 3. Cu trc la chn: switch..case  PAGEREF _Toc236050190 \h 30  HYPERLINK \l "_Toc236050191" 4. Cc cu trc lp  PAGEREF _Toc236050191 \h 33  HYPERLINK \l "_Toc236050192" 4.1. Cu trc lp while  PAGEREF _Toc236050192 \h 33  HYPERLINK \l "_Toc236050193" 4.2. Cu trc lp do..while  PAGEREF _Toc236050193 \h 35  HYPERLINK \l "_Toc236050194" 4.3. Cu trc lp for  PAGEREF _Toc236050194 \h 37  HYPERLINK \l "_Toc236050195" 5. Cu lnh break, continue, goto v hm exit  PAGEREF _Toc236050195 \h 39  HYPERLINK \l "_Toc236050196" 5.1. Cu lnh break  PAGEREF _Toc236050196 \h 39  HYPERLINK \l "_Toc236050197" 5.2. Cu lnh continue  PAGEREF _Toc236050197 \h 40  HYPERLINK \l "_Toc236050198" 5.3. Cu lnh goto  PAGEREF _Toc236050198 \h 40  HYPERLINK \l "_Toc236050199" 5.4. Hm exit  PAGEREF _Toc236050199 \h 41  HYPERLINK \l "_Toc236050200" Chng 5. Hm  PAGEREF _Toc236050200 \h 42  HYPERLINK \l "_Toc236050201" 1. Khi nim  PAGEREF _Toc236050201 \h 42  HYPERLINK \l "_Toc236050202" 2. Khai bo hm  PAGEREF _Toc236050202 \h 42  HYPERLINK \l "_Toc236050203" 2.1. C php khai bo nguyn mu hm  PAGEREF _Toc236050203 \h 42  HYPERLINK \l "_Toc236050204" 2.2. nh ngh)a hm  PAGEREF _Toc236050204 \h 42  HYPERLINK \l "_Toc236050205" 3. Kt qu tr v ca hm  Lnh return  PAGEREF _Toc236050205 \h 43  HYPERLINK \l "_Toc236050206" 4. Li gi hm v Cch truyn tham s cho hm  PAGEREF _Toc236050206 \h 44  HYPERLINK \l "_Toc236050207" 5.  qui  PAGEREF _Toc236050207 \h 45  HYPERLINK \l "_Toc236050208" 5.1. Khi nim  PAGEREF _Toc236050208 \h 45  HYPERLINK \l "_Toc236050209" 5.2. V d  PAGEREF _Toc236050209 \h 45  HYPERLINK \l "_Toc236050210" Bi tp luyn:  PAGEREF _Toc236050210 \h 45  HYPERLINK \l "_Toc236050211" Chng 6. Mng  PAGEREF _Toc236050211 \h 47  HYPERLINK \l "_Toc236050212" 1. Khi nim  PAGEREF _Toc236050212 \h 47  HYPERLINK \l "_Toc236050213" 2. Khai bo mng  PAGEREF _Toc236050213 \h 47  HYPERLINK \l "_Toc236050214" 2.1. Khai bo mng  PAGEREF _Toc236050214 \h 47  HYPERLINK \l "_Toc236050215" 2.2. Truy xut n cc phn t ca mng  PAGEREF _Toc236050215 \h 47  HYPERLINK \l "_Toc236050216" 3. Khi to mng  PAGEREF _Toc236050216 \h 48  HYPERLINK \l "_Toc236050217" 4. Dng mng lm tham s  PAGEREF _Toc236050217 \h 50  HYPERLINK \l "_Toc236050218" 5. Vi mng hai chiu  PAGEREF _Toc236050218 \h 52  HYPERLINK \l "_Toc236050219" 5.1. nh ngh)a  PAGEREF _Toc236050219 \h 52  HYPERLINK \l "_Toc236050220" 5.2. Truy xut cc phn t mng hai chiu  PAGEREF _Toc236050220 \h 52  HYPERLINK \l "_Toc236050221" 5.3. Khi to gi tr mng hai chiu  PAGEREF _Toc236050221 \h 53  HYPERLINK \l "_Toc236050222" 5.4. V d  PAGEREF _Toc236050222 \h 54  HYPERLINK \l "_Toc236050223" Bi tp luyn  PAGEREF _Toc236050223 \h 55  HYPERLINK \l "_Toc236050224" Chng 7. Con tr  PAGEREF _Toc236050224 \h 56  HYPERLINK \l "_Toc236050225" 1. Khi nim  PAGEREF _Toc236050225 \h 56  HYPERLINK \l "_Toc236050226" 2. Ton t ly a ch (&)  PAGEREF _Toc236050226 \h 56  HYPERLINK \l "_Toc236050227" 3. Ton t tham chiu (*)  PAGEREF _Toc236050227 \h 57  HYPERLINK \l "_Toc236050228" 4. Khai bo bin kiu con tr.  PAGEREF _Toc236050228 \h 57  HYPERLINK \l "_Toc236050229" 5. Cc php ton  PAGEREF _Toc236050229 \h 58  HYPERLINK \l "_Toc236050230" 5.1. Php gn  PAGEREF _Toc236050230 \h 58  HYPERLINK \l "_Toc236050231" 5.2. Php tng gim a ch  PAGEREF _Toc236050231 \h 58  HYPERLINK \l "_Toc236050232" 5.3. Php truy nhp b nh  PAGEREF _Toc236050232 \h 58  HYPERLINK \l "_Toc236050233" 5.4. Php so snh  PAGEREF _Toc236050233 \h 59  HYPERLINK \l "_Toc236050234" 6. Con tr hng  PAGEREF _Toc236050234 \h 59  HYPERLINK \l "_Toc236050235" 7. Con tr mng  PAGEREF _Toc236050235 \h 59  HYPERLINK \l "_Toc236050236" 8. Khi to con tr  PAGEREF _Toc236050236 \h 60  HYPERLINK \l "_Toc236050237" 9. Con tr tr ti con tr  PAGEREF _Toc236050237 \h 62  HYPERLINK \l "_Toc236050238" 10. Con tr khng kiu  PAGEREF _Toc236050238 \h 62  HYPERLINK \l "_Toc236050239" 11. Con tr hm  PAGEREF _Toc236050239 \h 63  HYPERLINK \l "_Toc236050240" Chng 8. Cu trc  PAGEREF _Toc236050240 \h 65  HYPERLINK \l "_Toc236050241" 1. Khi nim cu trc  PAGEREF _Toc236050241 \h 65  HYPERLINK \l "_Toc236050242" 2. Khai bo cu trc  PAGEREF _Toc236050242 \h 65  HYPERLINK \l "_Toc236050243" 2.1. Kiu cu trc  PAGEREF _Toc236050243 \h 65  HYPERLINK \l "_Toc236050244" 2.2. Khai bo thnh phn (bin, mng) kiu cu trc  PAGEREF _Toc236050244 \h 67  HYPERLINK \l "_Toc236050245" 3. Truy cp n cc thnh phn ca cu trc  PAGEREF _Toc236050245 \h 67  HYPERLINK \l "_Toc236050246" 4. V d cu trc  PAGEREF _Toc236050246 \h 68  HYPERLINK \l "_Toc236050247" Bi tp luyn:  PAGEREF _Toc236050247 \h 71  HYPERLINK \l "_Toc236050248" Chng 9. File  PAGEREF _Toc236050248 \h 73  HYPERLINK \l "_Toc236050249" 1. Khi nim File  PAGEREF _Toc236050249 \h 73  HYPERLINK \l "_Toc236050250" 2. To file c file  PAGEREF _Toc236050250 \h 73  HYPERLINK \l "_Toc236050251" 2.1. Khai bo con tr tr n tp tin  PAGEREF _Toc236050251 \h 74  HYPERLINK \l "_Toc236050252" 2.2. M tp tin  PAGEREF _Toc236050252 \h 74  HYPERLINK \l "_Toc236050253" 2.3. Cc kiu x l tp thng dng  PAGEREF _Toc236050253 \h 75  HYPERLINK \l "_Toc236050254" 2.4. ng tp tin  PAGEREF _Toc236050254 \h 76  HYPERLINK \l "_Toc236050255" 2.5. Kim tra n cui tp tin hay cha  PAGEREF _Toc236050255 \h 76  HYPERLINK \l "_Toc236050256" 2.6. Cc x l trn tp tin  PAGEREF _Toc236050256 \h 76  HYPERLINK \l "_Toc236050257" 2.7. Truy cp n tp tin vn bn (text)  PAGEREF _Toc236050257 \h 77  HYPERLINK \l "_Toc236050258" 3. To file nh phn  PAGEREF _Toc236050258 \h 80  HYPERLINK \l "_Toc236050259" 4. c file nh phn  PAGEREF _Toc236050259 \h 81  HYPERLINK \l "_Toc236050260" 4.1. Ghi d liu ln tp nh phn  PAGEREF _Toc236050260 \h 81  HYPERLINK \l "_Toc236050261" 4.2. c d liu t tp tin nh phn - Hm fread()  PAGEREF _Toc236050261 \h 81  HYPERLINK \l "_Toc236050262" 4.3. Di chuyn con tr tp tin - Hm fseek()  PAGEREF _Toc236050262 \h 81  HYPERLINK \l "_Toc236050263" Bi tp luyn  PAGEREF _Toc236050263 \h 83  HYPERLINK \l "_Toc236050264" Ti liu tham kho  PAGEREF _Toc236050264 \h 87  Chng 1. Thut ton 1. V d + Thut ton gii phng trnh bc 1, gii phng trnh bc 2. + Thut ton tm s ln nht trong 3 s a, b, c. + Thut ton tnh trung bnh cng ca 4 s a, b, c, d. + Thut ton tm ng i ngn nht. & 2. Khi nim Thut ng thut ton (Algorithm) l t vit tt ca tn mt nh ton hc  th k IX: Abu Ja fa Mohammed ibn Musa al-Khowarizmi. u tin, thut ton c hiu nh l cc quy tc thc hin cc php ton s hc vi cc con s c vit trong h thp phn. Cng vi s pht trin ca my tnh, khi nim thut ton c hiu theo ngh)a rng hn. Mt nh ngh)a hnh thc v thut ton c nh ton hc ngi Anh l Alanh Turing a ra vo nm 1936 thng qu my Turing. C th ni l thuyt thut ton c hnh thnh t . L thuyt thut ton lin quan n vn  sau: + Gii c bng thut ton: Thay nhng thut ton cha tt bng nhng thut ton tt hn. + Trin khai thut ton: Xy dng nhng ngn ng thc hin trn my tnh  m ha thut ton. Vy, Thut ton l mt dy hu hn cc thao tc c b tr theo mt trnh t xc nh, c  ra trc, nhm gii quyt mt bi ton nht nh. Thao tc hay cn gi l tc v, php ton (Operation) hay lnh (Command), ch th (Instruction)& l mt hnh ng cn c thc hin bi c ch thc hin thut ton. Mi thao tc bin i bi ton t mt trng thi trc (hay trng thi nhp) sang trng thi sau (hay trng thi xut). Thc t mi thao tc thng s dng mt s i tng trong trng thi nhp (cc i tng nhp) v sn sinh ra cc i tng mi trong trng thi xut (cc i tng xut). Quan h gia 2 trng thi xut v nhp cho thy tc ng ca thao tc. Dy cc thao tc ca thut ton ni tip nhau nhm bin i bi ton t trng thi ban u n trng thi kt qu. Mi thao tc c th phn tch thnh cc thao tc n gin hn. Trnh t thc hin cc thao tc phi c xc nh r rng trong thut ton. Cng mt tp hp thao tc nhng xp t theo trnh t khc nhau s cho kt qu khc nhau. 3. Cc c trng ca thut ton + Tnh xc nh: Cc thao tc, cc i tng, phng tin trong thut ton phi c ngh)a r rng, khng c gy nhm ln. Ni cch khc, hai c ch hot ng khc nhau cng thc hin mt thut ton, s dng cc i tng, phng tin nhp phi cho cng mt kt qu. + Tnh dng: i hi thut ton phi dng v cho kt qu sau mt s hu hn cc bc. + Tnh ng ca thut ton: Thut ton ng l thut ton cho kt qu tha mn c t thut ton vi mi trng hp ca cc i tng, phng tin nhp. + Tnh ph dng: Thut ton  gii mt lp bi ton gm nhiu bi c th, lp  c xc nh bi c t. D) nhin l c lp bi ton ch gm 1 bi. Thut ton khi  s khng cn s dng i tng, phng tin nhp no c. 4. Phng php biu din Thut ton c th din t di nhiu hnh thc, chng hn di dng lu , dng ngn ng t nhin, dng m gi hoc mt ngn ng lp trnh no khc. a. Dng ngn ng t nhin: Thut ton c th trnh by di dng ngn ng t nhin theo trnh t cc bc thc hin trong thut ton. b. Dng ngn ng lp trnh: Dng cu trc lnh, d liu ca mt ngn ng lp trnh no   m t. c. Dng m gi: Thut ton trnh by trong dng vn bn bng ngn ng t nhin tuy d hiu nhng kh ci t. Dng mt ngn ng lp trnh no   din t th phc tp, kh hiu. Thng thng thut ton cing c trao i di dng vn bn  tuy khng rng buc nhiu vo c php xc nh nh cc ngn ng lp trnh, nhng cing tun theo mt s quy c ban u  ta gi l dng m gi. Ty theo vic nh hng ci t thut ton theo ngn ng lp trnh no ta din t thut ton gn vi ngn ng y. d. Dng lu : Trong cc phng php biu din, chng ta s ch yu nghin cu phng php biu din theo dng ny. Dng lu  dng cc hnh v (c quy c)  din t thut ton. Lu  cho hnh nh trc quan v tng th ca thut ton, cho nn thng c s dng nhiu nht. Cc k hiu s dng trong phng php biu din thut ton bng lu : STTK hiuGii thch1Bt u v kt thc chng trnh2im ni, ng i (lung x l)3iu khin la chn4Thao tc nhp, xut.5Thao tc x l hoc tnh ton.6Tr v gi tr (return)7im ni lin kt tip theo (s dng khi lu  vt qu trang) V d 1: c cc thng tin nh tn, tui v lu li nhng ngi c tui trn 50. Lu :  V d 2: d: Nhp vo 3 s nguyn a, b, c v xut ra mn hnh vi gi tr ca mi s tng ln 1.  Ch khi v lu : + Trc tin hy tp trung v mt s ng i chnh ca lu . + Thm vo tt c cc nhnh v vng lp. + Mt lu  ch c mt im Bt u v mt im kt thc. + Mi bc trong chng trnh khng cn th hin trong lu . + Lu  cn phi p ng c yu cu: nhng ngi lp trnh khc c th hiu lu  mt cch d dng. Bi tp luyn V lu  thut ton cho cc bi ton sau: Bi 1: Tm s ln nht trong 3 s a, b, c. Bi 2: Gii phng trnh bc 1: ax + b = 0 (a `" 0). Bi 3: Gii phng trnh bc 2: ax2 + bx + c = 0 (a `" 0). Bi 4: Nhp vo hai s x, y. Xut ra mn hnh tng, hiu, tch, thng ca hai s trn. Bi 5: Nhp vo s nguyn n, kim tra xem n chn hay l v xut ra mn hnh. Bi 6: Nhp vo ba cnh a, b, c ca tam gic. Xut ra mn hnh tam gic  thuc loi tam gic g? (Thng, cn, vung, u hay vung cn). Bi 7: Tnh n!, vi n e" 0 Chng 2. Gii thiu ngn ng lp trnh C++ 1. Lch s hnh thnh Ngn ng lp trnh C do Dennis Ritchie ngh) ra khi ng lm vic ti AT&T Bell Laboratories vo nm 1972. Khng phi ngu nhin m Dennis Ritchie ngh) ra C. Vo thi gian  ng cng mt vi ng nghip c nhim v thit k mt h iu hnh mi m ngy nay chng ta bit n nh l h iu hnh UNIX. Trong qu trnh thit k, Dennis Ritchie nhn thy trong tt c cc ngn ng hin c lc , khng c ngn ng no thch hp  h c th s dng vo  n ca h c. Cui cng Dennis Ritchie quyt nh lm ra mt ngn ng mi  gii quyt nhng vn  m h gp phi. Dennis Ritchie  tham kho rt nhiu cc ngn ng c t trc, v c bit l ngn ng B do Ken Thompson cing lm vic ti Bell Labs ngh) ra. Rt nhiu phn ca ngn ng ny c ng a vo C. Chnh v vy m Dennis Ritchie  chn ngay ch ci tip theo ca B (tc l C)  t tn cho ngn ng mi ny. V C l mt ngn ng mnh v c tnh linh hot, n  nhanh chng c s dng mt cch rng ri, vt ra khi phm vi ca Bell Labs. Cc lp trnh vin  khp mi ni bt u s dng n  vit tt c cc loi chng trnh. Tuy nhin, nhiu t chc  nhanh chng a ra cc versions C ca ring h v nhng s khc bit tinh t ca cc phn c b sung bt u khin cc lp trnh vin au u. V vy vo nm 1983, Vin tiu chun Quc gia Hoa K (the American National Standards Institute - ANSI)  thnh lp mt y ban  a ra mt tiu chun cho C, c bit vi ci tn ANSI C. Tr mt vi ngoi l, tt c cc trnh dch C hin i u c gng theo st tiu chun ny. C++ c bit n nh l ngn ng mi bao trm ln C v do Bjarne Stroustrup sng tc nm 1980 cing ti phng th nghim Bell ti bang New Jersey, M. Ban u c ng t tn cho n l  C with classes (C vi cc lp). Tuy nhin n nm 1983 th ng i tn thnh C++, trong  ++ l ton t tng thm 1 ca C. Mc d C l mt ngn ng c gii lp trnh chuyn nghip yu thch song n vn c nhng hn ch ca n. Nu mt khi s dng lnh vt hn 25000 th cng vic s tr nn rt phc tp. C++ c to ra  xa b chng ngi ny. iu c bn ca C++ l cho php ngi lp trnh hiu v qun l cc chng trnh ln, phc tp hn. C++ c bit n nh l ngn ng lp trnh hng s vt hay hng i tng - OOP (Object Oriented Programming). 2. c im C l mt ngn ng mnh v linh hot.  Nhng g bn c th lm vi C ch thua c tr tng tng ca bn m thi . Ngn ng khng a ra cc rng buc i vi bn. C c s dng trong nhiu d n khc nhau, nh vit h iu hnh, chng trnh x l vn bn,  ho, bng tnh, v thm ch c chng trnh dch cho cc ngn ng khc. C l ngn ng c cc lp trnh vin chuyn nghip a thch hn c. Cing v vy m c sn rt nhiu cc trnh bin dch (compiler) v cc th vin c vit sn khc. C l mt ngn ng kh chuyn (portable language). gh)a l mt chng trnh vit bng C cho mt h my tnh (v d nh IBM PC) c th c dch v chy trn h my tnh khc (chng hn nh DEC VAX) ch vi rt t cc s i. Tnh kh chuyn  c bi chun ANSI cho C. C ch gm mt s t t kho (keywords) lm nn tng  xy dng cc cc chc nng ca ngn ng. C l bn ngh) rng mt ngn ng vi nhiu t kho (i khi cn c gi l t dnh ring - reserved words) s mnh hn. Khng phi nh vy. Khi lp trnh vi C, bn s thy rng n c th dng  lp trnh gii quyt bt k bi ton no. C l ngn ng lp trnh theo modul. M chng trnh C c th (v nn) c vit thnh cc th tc gi l function. Nhng function ny c th c s dng li trong cc ng dng (application) v chng trnh khc nhau. Tuy nhin C khng cho php khai bo hm trong hm.  l nhng c im ni bt ca C, v C++ bao trm ln C nn mi c im ca C u c trong C++. Ngoi ra, C++ cn c mt s c im khc nh: + C++ l ngn ng hng i tng. + C++ l ngn ng nh kiu rt mnh. + C++ cung cp cch truyn tham s bng tham chiu cho hm. + C++ cung cp c cu th vin  ngi lp trnh c th t to thm hm thng dng vo th vin v c th ti s dng sau ny. + C++ cung cp mt c ch a dng ha tn hm v ton t, cho php s dng cng mt tn hm hoc cng mt k hiu ca ton t  nh ngh)a cc chng trnh con thc hin cc nhim v tng t nhau vi tp kiu d liu tham s khc nhau. + C++ cung cp cc class l loi cu trc mi ng gi chung cho c d liu ln cc hm trong mt ch th c bo v mt cch cht ch. + C++ Cung cp cc class con trong  mt class c th k tha d liu v hm ca class khc l cc class cha m, to ra s di truyn. 3. Cu trc ca mt chng trnh C++ Mt chng trnh C++ ni chung c dng nh sau: (1): Khai bo th vin (2): [Khai bo cc nguyn mu hm ca ngi dng]. (3): [Cc nh ngh)a kiu]. (4): [Cc nh ngh)a Macro]. (5): [Cc nh ngh)a bin, hng]. (6): <kiu hm> main ([khai bo tham s]). (7): { (8): Thn hm main (9): } (10): Cc nh ngh)a hm ca ngi dng. Ch : Cc thnh phn trong cp ngoc vung [] c th c hoc khng trong chng trnh. Gii thch c php: (1): C php  khai bo th vin: #include<tn_th_vin.h> V d: #include<iostream.h>; #include<conio.h>;& (2): Cung cp tn hm, kiu hm, s i s v kiu ca tng i s ca hm. C php khai bo nguyn mu hm: <kiu hm> <tn hm> ([Khai bo cc i s]); V d: int chanle (int x); Trong , kiu hm l int, tn hm l chanle, i s l x v kiu ca i s l int. (3): nh ngh)a kiu mi: Ngoi nhng kiu chun  c cung cp sn ca ngn ng, ngi lp trnh c th nh ngh)a ra cc kiu mi t nhng kiu  c bng cch s dng t kha typedef. (4): nh ngh)a Macro: Khi nim macro l g? Gi s nh bn c mt ni dung (gi tr) no  v bn mun s dng n nhiu ln trong chng trnh, nhng bn khng mun vit trc tip n vo chng trnh lc bn son tho v mt vi l do no  (chng hn nh n s lm chng trnh kh c, kh hiu, hoc khi thay i s kh,..). Lc ny bn hy gn cho ni dung  mt  tn v bn s dng  tn   vit trong chng trnh ngun. Khi bin dch chng trnh, chng trnh dch s t ng thay th ni dung ca  tn vo ng v tr ca  tn . Thao tc ny gi l php th macro v chng ta gi  tn l tn ca macro v ni dung ca n c gi l ni dung ca macro. Mt macro c nh ngh)a nh sau: #define tn_macro ni_dung Trong  tn macro l mt tn hp l, ni dung (gi tr) ca macro c coi thun tu l 1 xu cn thay th vo v tr xut hin tn ca macro tng ng, gia tn v ni dung cch nhau 1 hay nhiu khong trng (du cch). Ni dung ca macro bt u t k t khc du trng u tin sau tn macro cho ti ht dng. (5): Cc nh ngh)a bin, hng: Cc bin v hng c nh ngh)a ti y s tr thnh bin v hng ton cc. ngh)a v bin, hng, c php nh ngh)a  c trnh by trong mc bin v hng. (6)  (9): Hm main():y l thnh phn bt buc trong mt chng trnh C++, thn ca hm main bt u t sau du m mc { (dng 7) cho ti du ng mc } (dng 9). (10): Cc nh ngh)a hm ca ngi dng: Mt nh ngh)a hm bao gm tiu  ca hm, thn hm vi c php nh sau: <Kiu hm> <Tn hm> ([Khai bo cc i]) { <Thn hm> } Ch : Tiu  trong nh ngh)a hm phi tng ng vi nguyn mu hm. Nu trong chng trnh nh ngh)a hm xut hin trc khi gp li gi hm  th c th khng nht thit phi c dng khai bo nguyn mu hm. 4. Mt s v d mu V d 1: Chng trnh nhp vo im 3 mn: Ton, L, Ha v tnh im trung bnh ca 3 mn . //Khai bo th vin #include<iostream.h> //Bt u hm main int main() { //Khai bo 3 bin double dtb,dt,dl,dh; //Nhp vo 3 im Ton, L, Ha cout<<"Diem toan la:"; cin>>dt; cout<<"diem ly la:"; cin>>dl; cout<<"Diem hoa la:"; cin>>dh; dtb = (dt+dl+dh)/3; cout<<"diem trung binh 3 mon la:"<<"="<=5&&dtb<7) cout<<"Hoc luc trung binh"; if (dtb>=7 && dtb<8) cout<<"Hoc luc kha"; else cout<<"Hoc luc gioi"; return 0; } //Kt thc hm main V d 2: Vit chng trnh c s dng hm tm s ln nht trong 3 s x, y, z; //Khai bo th vin #include<iostream.h> //Khai bo hm c tn l maxabc float maxabc (float a, float b, float c) { float max; max = a>b?a:b; return (max>c?max:c); } //Bt u hm main main() { //Khai bo 4 bin float x,y,z,t; //Nhp gi tr cho cc bin cout<<"x = "; cin>>x; cout<<"y = "; cin>>y; cout<<"z = "; cin>>z; t = maxabc (x,y,z); cout<<"So lon nhat trong 3 so la: "<<t; } //Kt thc hm main 5. Ci t chng trnh Chng ta s s dng Turbo C++ 4.5 trong phn ti liu ny.  ci t Turbo C++ 4.5 bn lm theo hng dn sau: Bc 1: Vo th mc cha b ci TurboC++ 4.5, chn file Install.exe, click p vo th mc . Xut hin hnh sau:  Bc 2: Click Continue. Mn hnh sau hin ra:  Bc 3: Tt mn hnh Install.txt. Mn hnh sau hin ra, bn click Continue:  Bc 4: Click Continue:  Bc 5: Mn hnh sau hin ra, bn click vo Install:  Bc 6: Qu trnh ci t bt u chy. Sau khi qu kt thc qu trnh , mn hnh sau hin ra, bn click Continue:  Bc 7: Mn hnh hin th Readme.txt. Bn close ca s  li. Click OK. Bc 8: Mn hnh hin th ng k:  Bc 9: Bn in y  thng tin vo cc mc trong hnh sau:  Bc 10: Click Continue/Ok. Cc bc ci t  hon tt. 6. Khi ng chng trnh  khi ng chng trnh bn s dng mt s cch sau: Cch 1: Start/Program/Turbo C++ 4.5/Turbo C++. Cch 2: Click p vo biu tng Turbo C++ trn mn hnh. 7. Son tho chng trnh Mn hnh son tho ca Turbo C++ 4.5 c dng sau:  8. Thot khi chng trnh Trc khi thot khi chng trnh Turbo C++, bn hy Save/Save as ton b chng trnh.  thot khi chng trnh c th s dng mt s cch sau: Cch 1: Click menu File/Exit. Cch 2: Click vo biu tng  bn phi mn hnh. Chng 3. Cc thnh phn v cc kiu d liu c bn 1. Cc thnh phn Cc thnh phn trong C++ bao gm: + T kha. + Cc khai bo bin, hng. + Cc cu trc iu khin. + Biu thc v cc php ton. + Cc nh ngh)a, khai bo hm, khai bo kiu,& 2. Cc kiu d liu cn bn Mt trong mc ch ca cc chng trnh l x l, bin i thng tin, cc thng tin cn x l phi c biu din theo mt cu trc xc nh no  ta gi l cc kiu d liu. Cc kiu d liu ny c quy nh bi ngn ng lp trnh, hay ni khc i mi ngn ng c tp cc kiu d liu khc nhau. Khng hon ton ging nh khi nim kiu d liu trong ton hc, trong cc ngn ng lp trnh ni chung mi kiu d liu ch biu din c mt min gi xc nh no . Chng hn nh s nguyn chng ta hiu l cc s nguyn t - " ti + ", nhng trong ngn ng lp trnh min cc gi tr ny b gii hn, s gii hn ny ph thuc vo kch thc ca vng nh biu din s . V vy khi ni ti mt kiu d liu chng ta phi  cp ti 3 thng tin c trng ca n  l: - Tn kiu d liu. - Kch thc vng nh biu din n,min gi tr. - Cc php ton c th s dng. Cc kiu d liu n gin trong C++ ch l cc kiu s, thuc hai nhm chnh  l s nguyn v s thc (s du phy ng). Nhm cc kiu nguyn gm c: char, unsigned char, int, unsigned int, short, unsigned short, long, unsigned long c m t trong bng sau: Kiu d liuTn kiu (t kho)Kch thcMin gi trK t c du char1 byte t -128 ti 127 K t khng duunsigned char1 bytet 0 ti 255 S nguyn c duint2 bytet -32768 ti 32767 S nguyn khng duunsigned int2 bytet 0 ti 65535 S nguyn ngn c du short2 bytet -32768 ti 32767 S nguyn ngn c du unsigned short2 byte t 0 ti 65535 S nguyn di c du long4 bytet -2,147,483,648 ti 2,147,438,647 S nguyn di khng du unsigned long4 bytet 0 ti 4,294,967,295  Nhm cc kiu s thc gm: float, double, long double Khun dng biu din ca s thc khng ging nh s nguyn. Mt s thc ni chung c biu din theo k php khoa hc gm phn nh tr v phn mi vit sau ch E  biu din s mi c s 10. V d: 3.14 = 314*10-2 s c vit theo k php khoa hc l: 314E  2. Hoc: 314 = 3.14 *102 s c vit theo k php khoa hc l: 3.14E + 2. Nhm cc kiu s thc c m t trong bng sau: Kiu d liuTn kiuKch thcMin gi trS thc vi  chnh xc n float 4 byte3.4e-38 -> 3.4e38 S thc vi  chnh xc kp double 8 byte1.7e-308 -> 1.7e308 S thc di vi  chnh xc kp long double 10 byte3.4e-4832 -> 1.1e 4932  3. Hng 3.1. Khi nim Hng l i lng c gi tr thuc mt kiu d liu nht nh, nhng gi tr ca hng khng th thay i trong thi gian tn ti ca n. C hai loi hng mt l cc hng khng c tn (chng ta s gi l hng thng)  l cc gi tr c th tc thi nh : 8, hay 9.5 hoc  d . Loi th hai l cc hng c tn ( gi l hng k hiu). Cc hng k hiu cing phi nh ngh)a trc khi s dng, tn ca hng c t theo quy tc ca tn. Sau y nu khng c iu g c bit th chng ta gi chung l hng. 3.2. nh ngh)a hng Cc hng c nh ngh)a bng t kho const vi c php nh sau: const <kiu_d_liu> <tn_hng> = <gi_tr>; hoc const <tn_hng> = <gi_tr>; Trong dng th hai, chng trnh dch t ng n nh kiu ca hng l kiu ngm nh, chng trnh dch s t ng chuyn kiu ca <gi_tr> v kiu int. V d: const int a = 5; // nh ngh)a hng a kiu nguyn, c gi tr l 5 const float x = 4; // hng x kiu thc, c gi tr l 4.0 const d = 7; // hng d kiu int, gi tr l 7 const c =  1 ; // hng c kiu int gi tr = 49 const char * s =  Ngon ngu C++ ;// s l hng con tr, tr ti xu  Ngo ngu C++ Cc hng s trong C++ c ngm hiu l h 10, nhng bn c th vit cc hng trong h 16 hoc 8 bng c php, gi tr s h 16 c bt u bng 0x, v d nh 0x24, 0xA, cc s h 8 bt u bi s 0, v d 025, 057. Cc hng k t c vit trong cp du nhy n   : v d  a ,  2 cc gi tr ny c C++ hiu l s nguyn c gi tr bng m ca k t;  a c gi tr l 97,  B c gi tr bng 66. Cc xu k t l dy cc k t c vit trong cp du nhy kp   , v d  Ngon ngu C ,  a (l kiu xu k t ). Ch : Cc bin, hng c th c nh ngh)a ngoi mi hm, trong hm hoc trong mt khi lnh. Vi C chun th khi nh ngh)a bin, hng trong mt khi th dng nh ngh)a phi  cc dng u tin ca khi, tc l trc tt c cc lnh khc ca khi, nhng trong C++ bn c th t dng nh ngh)a bt k v tr no. 4. Cc php ton ca C++ STTPHP TON NGH(AGHI CHPHP TON S HC1 + Cng 2 -Tr 3 * Nhn 4 / Chia ly phn nguyn 5 % Chia ly phn d PHP TON QUAN H1 > Ln hn 2 < Nh hn 3 >= Ln hn hoc bng 4 <= Nh hn hoc bng 5 = = Bng nhau 6 != Khc nhau PHP TON LOGIC1 ! NOT 2 && AND 3 || OR TON T TNG GIM1 ++ Tng 1 Nu ton t tng gim t trc th tng gim trc ri tnh biu thc hoc ngc li.2 --Gim 1 PHP TON THAO TC TRN BIT1 & AND 2 | OR 3 ^ XOR 4 << Dch tri 5 >> Dch phi 6 ~ Ly phn b theo bit  5. Xut nhp d liu Trong th vin iostream ca C++, cc thao tc vo ra c bn ca mt chng trnh c h tr bi hai dng d liu: cin  nhp d liu v cout  xut. Thng thng cout c gn vi mn hnh cn cin c gn vi bn phm. 5.1. Xut d liu (cout) Dng cout c s dng vi ton t  qu ti << (overloaded - bn s hiu r hn v thut ng ny trong phn lp trnh hng i tng). cout << "Output sentence"; // Hin th Output sentence ln mn hnh cout << 120; // Hin th s 120 ln mn hnh cout << x; // Hin th ni dung bin x ln mn hnh Ton t << c gi l ton t chn v n chn d liu i sau n vo dng d liu ng trc. Trong v d trn n chn chui "Output sentence", hng s 120 v bin x vo dng d liu ra cout. Ch rng  dng u tin chng ta s dng du ngoc kp v  l mt chui k t. Khi chng ta mun s dng cc hng xu k t ta phi t chng trong cp du ngoc kp  chng c th c phn bit vi cc bin. V d, hai lnh sau y l hon ton khc nhau: cout << "Hello"; // Hin th Hello ln mn hnh cout << Hello; // Hin th ni dung ca bin Hello ln mn hnh Ton t chn (<<) c th c s dng nhiu ln trong mt cu lnh: cout << "Hello, " << "I am " << "a C++ sentence"; Cu lnh trn s in thng bo Hello, I am a C++ sentence ln mn hnh. S tin li ca vic s dng lp li ton t chn (<<) th hin r khi chng ta mun hin th nhiu bin v hng hn l ch mt bin: cout << "Hello, I am " << age << " years old and my email address is " << email_add; Cn phi nhn mnh rng cout khng nhy xung dng sau khi xut d liu, v vy hai cu lnh sau: cout << "This is a sentence."; cout << "This is another sentence."; s c hin th trn mn hnh: This is a sentence.This is another sentence. Bi vy khi mun xung dng chng ta phi s dng k t xung dng, trong C++ l \n: cout << "First sentence.\n "; cout << "Second sentence.\nThird sentence."; s vit ra mn hnh nh sau: First sentence. Second sentence. Third sentence. Thm vo ,  xung dng bn c th s dng tham s endl. V d: cout << "First sentence." << endl; cout << "Second sentence." << endl; s in ra mn hnh: First sentence. Second sentence. Tham s endl c mt tc dng c bit khi n c dng vi cc dng d liu s dng b m: Cc b m s c flushed (chuyn ton b thng tin t b m ra dng d liu). Tuy nhin, theo mc nh cout khng s dng b m. Cc k t iu khin y l cc k t c bit, bt u bng k t \ tip theo l 1 k t dng  iu khin: chuyn con tr mn hnh, v tr in d liu,.. - \n : chuyn con tr mn hnh xung dng mi - \t : du tab - \b : (backspace) li mt k t (xo k t trc v tr con tr hin ti). 5.2. Nhp d liu (cin) Thao tc vo chun trong C++ c thc hin bng cch s dng ton t  qu ti >> vi dng cin. Theo sau ton t ny l bin s lu tr d liu c c vo. V d: int age; cin >> age; khai bo bin age c kiu int v i nhp d liu t cin (bn phm)  lu tr n trong bin kiu nguyn ny. cin ch bt u x l d liu nhp t bn phm sau khi phm Enter c g. V vy d bn ch nhp mt k t th cin vn s kin nhn ch cho n khi bn g phm Enter. Chng 4. Cc cu trc iu khin 1. Lnh n v lnh phc 1.1. Lnh n Mt biu thc kiu nh: a =5; a+ =6;& tr thnh cu lnh khi c i km theo du chm phy (;) pha cui cu. V d: a = 5; a+=6; & 1.2. Lnh phc hay khi lnh Lnh phc l mt dy cc cc cu lnh c bao bi cp du ngoc nhn {}. V d 1: { S+=(i*i); i++; } V d 2: { cout<<a[i][j]<<" "; if (j==i) cout<<"\n"; } 2. Cu trc iu kin if..else 2.1. Dng khuyt If (biu thc iu kin) { Khi lnh; } Cch hot ng: u tin, chng trnh kim tra biu thc iu kin sau if. Nu biu thc iu kin ng th cc cu lnh trong khi lnh trong cu trc if c thc hin. Ngc li, chng trnh s thot ra khi cu trc if v thc hin cc lnh tip theo sau cu trc if. Biu din di dng lu  cch hot ng ca cu trc if:  2.2. Dng y  if (biu thc iu kin) { Khi lnh 1; } else { Khi lnh 2; } Cch hot ng: u tin, chng trnh kim tra biu thc iu kin sau if, nu biu thc iu kin ng th thc hin cc cu lnh trong khi lnh 1. Ngc li, nu biu thc iu kin sai s thc hin cc cu lnh trong khi lnh 2. Biu din bng lu  ca cu trc if..else:  V d 1: S dng cu trc if v if..else  nhp vo im ca 3 mn: Ton, l, ha v tnh im trung bnh: #include<iostream.h> int main() { double dtb,dt,dl,dh; cout<<"Diem toan la:"; cin>>dt; cout<<"diem ly la:"; cin>>dl; cout<<"Diem hoa la:"; cin>>dh; dtb = (dt+dl+dh)/3; cout<<"diem trung binh 3 mon la:"<<"="<=5&&dtb<7) cout<<"Hoc luc trung binh"; if (dtb>=7 && dtb<8) cout<<"Hoc luc kha"; else cout<<"Hoc luc gioi"; return 0; } V d 2: Gii phng trnh bc 1: ax + b = 0 #include<iostream.h> #include<math.h> int main() { int a,b; float x; cout<<"Nhap a va b:"<<"\n"; cin>>a>>b; if (a==0 && b==0) cout<<"Bat phuong trinh co vo so nghiem"; if (a==0 && b!=0) cout<<"Bat phuong trinh vo nghiem"; if (a!=0) cout<<"Bat phuong trinh co nghiem x = "<<-b/a; } V d 3: Vit chng trnh nhp vo 1 s, tnh cn bc 2 ca s  //Chuong trinh tinh can bac 2 #include<iostream.h> #include<math.h> main() { int a; cout<<"Nhap a = "; cin>>a; if (a<0) cout<<"Khong tinh duoc"; if (a>0) cout<<"Can bac hai cua a la: "<< sqrt(a); } 3. Cu trc la chn: switch..case Cu trc switch..case cho php la chn mt nhnh trong nhiu kh nng ty vo iu kin xy ra. C php: switch (Biu thc nguyn) { case n1: Khi lnh 1; case n2: Khi lnh 2; & case nk: Khi lnh k; [default: Khi lnh; ] } Trong : ni l cc s nguyn, hng k t hoc biu thc hng. Cc ni cn c gi tr khc nhau. on chng trnh nm trong cp ngoc nhn {} l thn ca switch. default, case l cc t kha. default l thnh phn khng bt buc. Cch hot ng: s hot ng ca cu trc switch..case ph thuc vo gi tr ca biu thc nguyn sau switch. u tin <biu thc nguyn> c tnh, sau  ln lt so snh gi tr ca <bt nguyn> vi cc biu thc hng <n1>, <n2>,..<nk>. Nu gi tr ca mt biu thc hng th k trong cc biu thc ny trng vi gi tr ca <biu thc nguyn> th chng trnh s thc hin cc lnh bt u t khi lnh k v tip tc cc lnh pha di cho ti khi: + Gp cu lnh : break (tt nhin nu gp cc cu lnh return, exit th cing kt thc). + Gp du ng mc } ht cu trc switch. Nu <biu thc nguyn> khng trng vi gi tr no trong cc biu thc hng th khi lnh (cc lnh sau mnh  default nu c) s c thc hin, ri ra khi cu trc switch. V d 1: S dng cu trc switch..case, hy vit chng trnh: nu nhp vo l  P hoc  p th hin th dng ch  Cao dang co dien Ha Noi . #include<iostream.h> main() { char ch; cout<<"Nhap vao ky tu P hoac p: "; cin>>ch; switch (ch) { case 'P': case 'p': cout<<"Cao dang co dien Ha Noi"; } } V d 2: Vit chng trnh s dng cu trc switch..case, nhp 2 s nguyn x, y t bn phm, chng trnh a ra la chn: + Nu ngi s dng nhp  + : Thc hin: x + y; + Nu ngi s dng nhp  - : Thc hin: x  y; + Nu ngi s dng nhp  * : Thc hin: x * y; + Nu ngi s dng nhp  / : Thc hin: x / y; Nu ngi s dng khng nhp mt trong cc ton t trn th a ra dng thng bo:  Khong hieu toan tu nay!. #include int main() { int X; cout<< "Nhap vao so thu nhat:"; cin>>X; int Y; cout<< "Nhap vao so thu hai:"; cin>>Y; char Op; cout<<"Nhap vao toan tu (+-*/):"; cin>>Op; switch(Op) { case '+': cout<<"Ket qua:"< Bc 2: nu gi tr tnh c ca <biu thc> l  sai (==0) th kt thc vng lp while. Bc 3: nu gi tr ca <biu thc> l  ng (!=0) th thc hin khi lnh sau while. Bc 4: quay li bc 1 Ch : Thn while c th c thc hin mt ln hoc nhiu ln v cing c th khng c thc hin ln no nu ngay t u biu thc sau while  sai. Biu din bng lu :  V d 1: #include<iostream.h> main() { int n; cout<< "Nhap vao 1 so (hay day so khong cach)"; cin>>n; while (n>0) { cout<< n<< ","; --n; } cout<< "Ket thuc"; return 0; } V d 2: Vit chng trnh tnh v in S = 1 +1/(2!) +1/(3!)+..+1/(n!) vi n nhp t bn phm, 0<n<10. #include const int Max =10; // gioi han void main() { int n, k ; float S, pt; cout<<"\nNhap n = "; cin>>n; while((n<=0)||(n>Max)) { cout<<"\nNhap lai n (0>n; } S=pt=1; k=2; while(k<=n) { pt /=k; S+=pt; k++; } cout<<"\nGia tri tong S = <0, tnh v in gi tr n! (giai tha ca n), vi n! = 1*2*& *(n-1)*n #include <iostream.h> const int Max =10; // gii hn gi tr cn tnh void main() { int n, i; long gt; cout<<"\nNhap n = "; cin>>n; while((n<=0)||(n>Max)) { cout<<"\nNhap lai n (0<n<= : "<< Max; cin>>n; } gt=i=1; while(i<=n) { gt*=i; i++; } cout<<"\nGia tri n =  <<gt; } 4.2. Cu trc lp do..while Trong ton t while vic kim tra iu kin kt thc t  u chu trnh. Khc vi while, do..while vic kim tra iu kin kt thc t cui chu trnh. Nh vy thn ca chu trnh bao gi cing c thc hin t nht mt ln. C php: do Khi lnh; while (biu thc); S hot ng ca vng lp do..while: Bc 1: thc hin khi lnh sau do. Bc 2: kim tra gi tr biu thc <biu thc> sau while, nu c gi tr  ng ( khc 0) th lp li bc 1, nu  sai (=0) th kt thc vng lp. Biu din di dng lu :  V d 1: #include<iostream.h> main() { unsigned long n; do { cout<< "Nhap vao 1 day so khong cach"; cin>>n; cout<< "Ban da nhap so"<<n<<endl; } while (n!=0); return 0; } V d 2: #include<iostream.h> int main() { unsigned long n; do { cout<<"Nhap vao 1 day so:"; cin>>n; cout<<" Ban vua nhap day so la:"<<n<< "\n"; } while (n!=0); return 0; } 4.3. Cu trc lp for C php: for (biu thc 1; biu thc 2; biu thc 3) Khi lnh; Trong : Biu thc 1: biu thc khi u. Biu thc 2: biu thc iu kin - iu kin lp. Biu thc 3: bc nhy - thng dng vi ngh)a l thay i bc nhy. C 3 biu thc ny u l tu chn, chng c th vng mt trong cu lnh c th nhng cc du chm phy vn phi c. S hot ng ca vng lp for: Hot ng ca for theo cc bc sau: Bc 1: Thc hin biu thc khi u  Biu thc 1. Bc 2: Tnh gi tr biu thc 2  xc nh iu kin lp. Nu biu thc 2 c gi tr  sai (==0) th ra khi vng lp. Ngc li, nu biu thc c gi tr  ng ( khc 0) th chuyn ti bc 3. Bc 3: Thc hin khi lnh sau for ( thn ca for ), chuyn ti bc 4. Bc 4: Thc hin biu thc, ri quay v bc 2. Biu din di dng lu :  V d 1: #include<iostream.h> main() { for (int n = 10; n>0; n--) { cout<< n<< ","; } cout<< "Ket thuc"; return 0; } V d 2: Nhp vo mt dy s nguyn, tm s ln nht v nh nht trong dy. #include <iostream.h> void main() { int n,a,max,min,i; do { cout<<"Nhap so phan tu cua day : "; cin>>n; }while(n<1); cout<<"Nhap so thu nhat : "; cin>>a; max=min=a; for(i=2; i<=n; i++) { cout<<"Nhap so thu nhat : "; cin>>a; if(a>max) max=a; else if(a int main () { int n=10; loop: ; cout << n << ", "; n--; if (n>0) goto loop; cout << "Kt thc!"; return 0; } 5.4. Hm exit Hm exit() trong C/C++ c s dng  thot khi chng trnh. Hm ny, khi c triu gi s ngay lp tc kt thc chng trnh v chuyn quyn iu khin cho h iu hnh. C php: exit (int m_tr_v); m_tr_v thng l s 0. S 0 s xc nh vic kt thc chng trnh mt cch bnh thng. Tuy nhin c mt vi trng hp m_tr_v l nhng s khc 0  xc nh mt vi loi li. Bi tp luyn Bi 1: Vit chng trnh gii phng trnh bc nht ax + b = 0 Bi 2: Vit chng trnh gii phng trnh bc hai ax2 + bx + c = 0 (a `" 0) Bi 3: Vit chng trnh gii phng trnh bc hai ax2 + bx + c = 0 (a bt k) Bi 4: Vit chng trnh nhp vo N, kim tra N c phi l s nguyn t khng ? Bi 5: Vit chng trnh nhp vo N, tnh tng v in ra tt c cc s c s ca N. V d: N = 20 - Tp cc c s ca 20 l: 1, 2, 4, 5, 10, 20 - Tng cc c s = 42 Bi 6: Vit chng trnh nhp vo s nguyn dng N, tnh tng a. S1 = 1 + 2 + 3 + 4 +& ..+ N b. S2 = 12 + 22 + 32 + 42 +& ..+ N2 Chng 5. Hm 1. Khi nim Hm l mt on chng trnh c lp thc hin trn vn mt cng vic nht nh sau  tr v gi tr cho chng trnh gi n, hay ni cch khc hm l s chia nh ca chng trnh. 2. Khai bo hm 2.1. C php khai bo nguyn mu hm <Kiu_hm> Tn_hm ([Danh_sch_tham_s]); Trong : Tn_hm: l mt tn hp l theo quy tc v tn ca ngn ng C/C++. Mi hm c tn duy nht v khng c trng vi cc t kha. Tn hm s c dng  gi hm. Kiu_hm: Hm c th tr v mt gi tr cho ni gi, gi tr  thuc mt kiu d liu no , kiu  c gi l kiu hm. Kiu hm c th l kiu chun cing c th l kiu do ngi dng nh ngh)a. Nu hm khng tr v gi tr th kiu hm l void. Danh_sch_tham_s: Hm c th nhn d liu vo thng qua cc tham s ca n (tham s hnh thc), cc tham s ny cing thuc kiu d liu xc nh. C th c nhiu tham s, cc tham s cch nhau bi du phy (,). Trong nguyn mu khng bt buc phi c tn tham s nhng kiu ca n th bt buc. Nu hm khng c tham s chng ta c th  trng phn ny hoc c th khai bo l void. V d: int max(int a, int b); // khai bo nguyn mu hm max, c hai tham s kiu int, kt qu tr v kiu int. float f(float, int); // nguyn mu hm f, c hai tham s, tham s th nht kiu float, tham s th 2 kiu int, kt qu tr v kiu float. void nhapmang(int a[], int ); // hm nhapmang, kiu void (khng c gi tr tr v), tham s th nht l mt mng nguyn, tham s th 2 l mt s nguyn. void g(); // hm g khng i, khng kiu. 2.2. nh ngh)a hm C php: <kiu_hm> <tn_hm>([khai_bo_tham_s]) { < thn hm> } Dng th nht l tiu  hm (dng tiu ) cha cc thng tin v hm: tn hm, kiu ca hm (hai thnh phn ny ging nh trong nguyn mu hm) v khai bo cc tham s (tn v kiu) ca hm, nu c nhiu hn mt th cc tham s cch nhau bi du phy(,). Thn hm l cc lnh nm trong cp { }, y l cc lnh thc hin chc nng ca hm. Trong hm c th c cc nh ngh)a bin, hng hoc kiu d liu; cc thnh phn ny tr thnh cc thnh phn cc b ca hm. V d: unsigned long giaithua (int n) { unsigned long ketqua =1; int i; for (i =2; i<=n; i++) ketqua = ketqua*i; return ketqua; } 3. Kt qu tr v ca hm  Lnh return Nu hm c gi tr tr v (kiu hm khc void) th trong thn hm trc khi kt thc phi c cu lnh tr v gi tr: return <gi tr>; <gi_tr> sau lnh return chnh l gi tr tr v ca hm, n phi c kiu ph hp vi kiu ca hm c khai bo trong dng tiu . Trng hp hm void chng ta c th dng cu lnh return (khng c gi tr)  kt thc hm hoc khi thc hin xong lnh cui cng (gp } cui cng) hm cing kt thc. V d: Hm nhp mt mng c n phn t nguyn: Tn hm: nhapmang. Gi tr tr v: khng tr v. Tham s: c hai tham s l mng cn nhp A v s phn t cn nhp N Nguyn mu hm l: void nhapmang (int [], int); nh ngh)a hm nh sau: void nhapmang (int A[], int N) { int i; cout<<"\nNhap so phan tu cua mang "; for(i=0;i<N; i++) { cout<<"a[ <<i<<  ]= "; cin>>a[i]; } return ; } 4. Li gi hm v Cch truyn tham s cho hm Mt hm c th gi thc hin thng qua tn hm, vi nhng hm c tham s th trong li gi phi truyn cho hm cc tham s thc s (i s) tng ng vi cc tham s hnh thc. Khi hm c gi v truyn tham s ph hp th cc lnh trong thn hm c thc hin bt u t lnh u tin sau du m mc { v kt thc khi gp lnh return, exit hay gp du ng mc } kt thc hm. C php: <tn_hm> ([danh sch cc tham s thc s]); Cc tham s thc s phi ph hp vi cc tham s hnh thc: S tham s thc s phi bng s tham s hnh thc. Tham s thc s c truyn cho cc tham s hnh thc tun t t tri sang phi, tham s thc s th nht truyn cho tham s hnh thc th nht, tham s thc s th 2 truyn cho tham s hnh thc th 2,.. kiu ca cc tham s hnh thc phi ph hp vi kiu ca cc tham s hnh thc. S ph hp  y c hiu l kiu trng nhau hoc kiu ca tham s thc s c th p v kiu ca tham s hnh thc. Vi C/C++ vic truyn tham s cho hm c thc hin qua c ch truyn tham tr. 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Hay ni khc i, cc tham s hnh thc l cc bin cc b trong hm, s thay i ca n trong hm khng nh hng ti cc bin bn ngoi. 5.  qui 5.1. Khi nim Mt hm c gi c tnh  qui nu trong thn ca hm  c lnh gi li chnh n mt cch tng minh hay tim n. 5.2. V d V d 1: Tnh tng S = 1 + 2 + 3 + & + n (n l s nguyn dng, c nhp t bn phm). Thut ton gii bng phng php  quy nh sau: long TongS (int n) { if(n==0) return 0; return (TongS(n-1) + n); } V d 2: Tnh gi tr ca n! (vi n c nhp t bn phm) Thut ton gii bng phng php  quy nh sau: long GiaiThua (int n) { if(n==0) return 1; return (GiaiThua(n-1) * n); } Bi tp luyn: Bi 1: Vit chng trnh tnh din tch v chu vi ca hnh ch nht vi chiu di v chiu rng c nhp t bn phm. Bi 2: Vit chng trnh tnh din tch v chu vi hnh trn vi bn knh c nhp t bn phm. Bi 3: Nhp s nguyn dng n (n>0). Lit k tt c cc s nguyn t nh hn n. Bi 4: Vit chng trnh tnh tin lng ngy cho cng nhn, cho bit trc gi vo ca, gi ra ca ca mi ngi. Gi s rng: - Tin tr cho mi gi trc 12 gi l 6000 v sau 12 gi l 7500. - Gi vo ca sm nht l 6 gi sng v gi ra ca tr nht l 18 gi (Gi s gi nhp vo nguyn). Bi 5: Nhp vo 3 s thc a, b, c v kim tra xem chng c thnh lp thnh 3 cnh ca mt tam gic hay khng? Nu c hy tnh din tch, chiu di mi ng cao ca tam gic v in kt qu ra mn hnh. Bit rng: - Cng thc tnh din tch s = sqrt(p*(p-a)*(p-b)*(p-c) ) - Cng thc tnh cc ng cao: ha = 2s/a, hb=2s/b, hc=2s/c. (Vi p l na chu vi ca tam gic). Bi 6: Nhp vo 6 s thc a, b, c, d, e, f . Gii h phng trnh sau : ax + by = c dx + ey = f Bi 7: Vit chng trnh nhp 2 s nguyn dng a, b. Tm USCLN v BSCNN ca hai s nguyn . Bi 8: Vit chng trnh tnh tng nghch o ca n giai tha. Bi 9: Vit chng trnh nhp s nguyn dng n gm k ch s (0<k<=5), tnh tng cc c s dng ca n. V d: Nhp n=6 Tng cc c s t 1 n n: 1+2+3+6=12.  Chng 6. Mng 1. Khi nim Mng thc cht l mt bin c cp pht b nh lin tc v bao gm nhiu bin thnh phn. Cc thnh phn ca mng l tp hp cc bin c cng kiu d liu v cng tn. Do   truy xut cc bin thnh phn, ta dng c ch ch mc (ch s). 2. Khai bo mng 2.1. Khai bo mng C php: < Kiu_mng> < Tn_mng > [ < S_phn_t_ti_a> ] ; Trong : Kiu_mng: y l kiu ca mng, l tn mt kiu d liu  tn ti, c th l kiu chun hoc kiu d liu do ngi lp trnh nh ngh)a . Tn_mng: l tn ca mng, do ngi lp trnh t, theo quy tc v tn ca C/C++. S_phn_t: l hng (hoc biu thc hng) nguyn, dng l s phn t ca mng. V d: int a[10]; // Khai bo mng a, c s phn t ti a l 10, kiu d liu ca mng a l int. float MT[20]; //Khai bo mng MT, c s phn t ti a l 20, kiu d liu ca mng MT l float. 2.2. Truy xut n cc phn t ca mng C php : tn_mng [ch_s] v d a[1], MT[3]; ch_s l s th t ca phn t trong mng, cc phn t ca mng c nh ch s bt u t 0. Vi mng c n phn t th cc phn t ca n c ch s l 0, 1,..,n-1. 3. Khi to mng Cc phn t ca mng cing nh cc bin n, chng ta c th khi to gi tr ban u cho chng trn dng nh ngh)a mng (gi l khi u) vi c php sau: Kiu_mng tn_mng [ s_phn_t ] = {gt_0, gt_1,..,gt_k}; hoc Kiu_mng tn_mng [ ] = {gt_0, gt_1,..,gt_k}; Trong  cc thnh phn Kiu_mng , tn_mng, s_phn_t nh trong phn khai bo mng. gt_0, gt_1,.., gt_k l cc gi tr khi u (gi l b khi u) cho cc phn t tng ng ca mng, tc l gn tun t cc gi tr trong b khi u cho cc phn t ca mng t tri qua phi. Trong dng th nht, s gi tr trong bn khi u ch c th <= s phn t ca mng (k d" s_phn_t). Khi  nhng phn t mng tha ra (khng c gi tr khi u) s t ng c gn bng 0 (trong trng hp mng s, nu l con tr s l NULL (rng) ). V d: int a[3] ={ 1, 3, 4}; th gi tr ca a[0] l 1, a[1] l 3, a[2] l 4. int b[5] ={1,2}; th gi tr ca b[0] l 1, b[1] l 2, b[3]=b[4] l 0. vi mng cc k t hoc xu k t th c hai cch khi u nh sau: char c[4] ={ a , b , c }; // c[0] l  a , c[1] l  b , c[2] l  c , c[3] l  \0 char s[10] = ABC ; // tng ng vi char s[10] ={ A , B , C , \0 } (nu s gi tr trong b khi u > s phn t mng chng trnh dch s bo li) Trong dng th hai, chng ta khng xc nh s phn t ca mng, trong trng hp ny chng trnh bin dch s t ng xc nh kch thc (s phn t) ca mng theo s gi tr trong b khi u. V d: int a[] ={1,3,4}; th a l mng c 3 phn t, gi tr ca a[0] l 1, a[1] l 3, a[2] l 4. V d 1: Vit chng trnh nhp vo mt mng A c n phn t kiu nguyn, n<=20, n c nhp t bn phm. In cc phn t ca mng theo trt t xui, ngc (theo trt t ch s). #include<iostream.h> main() { const max = 20; int n, i, a[max]; do { cout<<"Nhap vao so phan tu cua mang: "; cin>>n; } while (n < 1 || n > max); cout<<"Nhap vao mang a: "<<endl; for (i = 0; i<n; i++) cin>>a[i]; cout<<"Cac phan tu mang theo TT xuoi la: "; for (i = 0; i<n; i++) cout<<a[i]<<"\t"; cout<<endl; cout<<"Cac phan tu mang theo TT nguoc la: "; for (i = n-1;i>=0; i--) cout<<a[i]<<"\t"; return 0; } V d 2: Nhp vo 2 mng A v B, c s phn t ti a l 10. Tnh tng ca A v B. //Tinh tong cua C = A + B #include<iostream.h> main() { const max = 10; int n, i, a[max], b[max], c[max]; do { cout<<"Nhap vao so phan tu cua mang: "; cin>>n; } while (n < 1 || n > max); cout<<"Nhap vao so PT mang a: "<<endl; for (i = 0; i<n; i++) cin>>a[i]; cout<<"Nhap vao so PT mang b: "<<endl; for (i = 0; i<n; i++) cin>>b[i]; cout<<"Tong C = A + B = "; for (i = 0; i< n; i++) c[i] = a[i] + b[i]; for (i = 0; i<n; i++) cout<<c[i]<<"\t"; return 0; } 4. Dng mng lm tham s Vo mt lc no  c th chng ta cn phi truyn mt mng ti mt hm nh l mt tham s. Trong C++, vic truyn theo tham s gi tr mt khi nh l khng hp l, ngay c khi n c t chc thnh mt mng. Tuy nhin chng ta li c php truyn a ch ca n, vic ny cing to ra kt qu thc t ging thao tc  trn nhng li nhanh hn nhiu v hiu qu hn.  c th nhn mng l tham s th iu duy nht chng ta phi lm khi khai bo hm l ch nh trong phn tham s kiu d liu c bn ca mng, tn mng v cp ngoc vung trng. V d, hm sau: void procedure (int arg[]) nhn vo mt tham s c kiu "mng ca char" v c tn arg.  truyn tham s cho hm ny mt mng c khai bo: int myarray [40]; ch cn gi hm nh sau: procedure (myarray); V d 1: #include <iostream.h> void printarray (int arg[], int length) { for (int n=0; n<length; n++) cout << arg[n] << " "; cout << "\n"; } int main () { int firstarray[] = {5, 10, 15}; int secondarray[] = {2, 4, 6, 8, 10}; printarray (firstarray,3); printarray (secondarray,5); return 0; } V d 2: Tnh tng C = A + B #include <iostream.h> void nhapmang(int a[], int n); void inmang(int a[], int n); void tong(int A[],int B[],int C[],int n); void main(){ const int max = 20; int A[max], B[max],C[max]; int n; do{ cout<<"\nNhap so phan tu mang = "; cin>>n; } while(n<1 || n>max); cout<<"\nNhap A \n"; nhapmang(A,n); cout<<"\nNhap B \n"; nhapmang(B,n); tong(A,B,C,n); cout<<"\nmang A: "; inmang(A,n); cout<<"\nmang B: "; inmang(B,n); cout<<"\nmang C: "; inmang(C,n); } void nhapmang(int a[], int n){ int i; cout<<"\nNhap mang co so phan tu: "<<n; for(i=0; i<n; i++) { cout<<"\nPhan tu thu"<<"["<<i<<"]"; cin>>a[i]; } } void inmang(int a[], int n){ int i; for(i=0; i<n; i++) cout<<a[i]; } void tong(int A[],int B[],int C[],int n){ int i; for (i = 0; i<n; i++) C[i]=A[i]+B[i]; } 5. Vi mng hai chiu 5.1. nh ngh)a Mng hai chiu c th hiu nh bng gm cc dng cc ct, cc phn t thuc cng mt kiu d liu no . Mng hai chiu c nh ngh)a nh sau. C php: Kiu_mng tn_mng [sd][sc]; Trong : - Kiu_mng: y l kiu ca mng, l tn mt kiu d liu  tn ti, c th l kiu chun hoc kiu d liu do ngi lp trnh nh ngh)a. - tn_mng: l tn ca mng, do ngi lp trnh t, theo quy tc v tn ca C/C++. - sd, sc: l hng (hoc biu thc hng) nguyn, dng tng ng l s dng v s ct mng, s phn t ca mng s l sd*sc. V d: int a[2][5]; // a l mng s nguyn c 2 dng, 5 ct (c 10 phn t). float D[3][10]; // D l mng s thc c 3 dng, 10 ct (c 30 phn t). char DS[5][30]; // DS l mng k t c 5 dng, 30 ct. 5.2. Truy xut cc phn t mng hai chiu Mt phn t ca mng 2 chiu c xc nh qua tn (tn ca mng) v ch s dng, ch s ct ca n trong mng theo c php sau: tn_mng [csd][csc] Vi csd l s nguyn xc nh ch s dng v csc l s hiu ct cing nh trong mng 1 chiu cc ch s c tnh t 0. Tc l 0 d"csd d"sd-1 v 0d"cscd"sc-1. 5.3. Khi to gi tr mng hai chiu Cc phn t mng hai chiu cing c th c khi u gi tr theo c php (4 dng sau): + Kiu_mng tn_mng [sd][sc]={{k_dng_1},{ k_dng_2},..,{ k_dng_k}}; + Kiu_mng tn_mng [ ][sc] = {{k_dng_1},{ k_dng_2},..,{ k_dng_k}}; + Kiu_mng tn_mng [sd][sc] = { gt_1, gt_2,...,gt_n }; + Kiu_mng tn_mng [ ][sc] = { gt_1, gt_2,...,gt_n }; C php trn c th gii thch nh sau: Dng 1: c k b gi tr s c gn cho k dng u tin ca mng (k d" sd ), vi mi dng (c coi nh mng mt chiu) c khi to gi tr nh mng mt chiu: dng th nht c khi u bi {k_dng_1}, dng th hai c khi u bi {k_dng_1},.., dng th k c khi u bi {k_dng_k}. Yu cu k d" sd, ngc li chng trnh s bo li. Cc dng cui ca mng nu khng c b khi u tng ng th s c t ng gn gi tr 0 (hoc NULL nu l con tr). Dng 2: (khng xc nh s dng) chng trnh dch s t ng n nh s dng ca mng bng s b khi u ( = k), sau  thc hin khi u nh dng 1. Dng 3: n gi tr trong b khi u c gn cho cc phn t mng theo cch: sc gi tr u tin trong cc gi tr khi u (gt_1,..,gt_sc) c gn tun t cho cc phn t ca dng th nht trong mng, sc phn t k tip s gn cho cc phn t  dng th 2,... nu phn t no ca mng khng c gi tr khi u s c gn 0 (con tr l NULL) - vi iu kin n d" sd*sc, ngc li l li. Dng 4: s dng ca mng s c chng trnh t tnh theo s gi tr trong b khi u theo cng thc sd = (n/sc) +1, v khi u nh dng 3. V d 1: int a[3][2] = {{1,2},{3},{4,5}}; th cc phn t ca a nh sau: a[0][0]=1, a[0][1]=2, a[1][0]=3, a[1][1]= 0, a[2][0]=4,a[2][1]=5; V d 2: int b[ ][2] = {{1,2},{3},{4,5}}; th l mng 3 dng, 2 ct cc phn t ca b nh sau: b[0][0]=1, b[0][1]=2, b[1][0]=3,b[1][1]= 0, b[2][0]=4,b[2][1]=5; V d 3: int c[ ][2] = {1,2,3,4,5}; th s dng ca c l mng 5/2 +1 =3 dng, cc phn t ca a nh sau: c[0][0]=1, c[0][1]=2, c[1][0]=3,c[1][1]= 4, b[2][0]=5,b[2][1]=0; 5.4. V d Vit chng trnh nhp vo mng A(n, m) vi 1<=n,m<=5 cc s nguyn. Yu cu: + In mng A ra mn hnh. + In ra mn hnh cc phn t chn ca mng. + In ra mn hnh cc phn t l ca mng. #include<iostream.h> main() { const max = 5;//Kich thuoc toi da int a[max][max]; int n,m,i,j; do { cout<<"Nhap so dong cua mang = ";cin>>n; cout<<"Nhap so cot cua mang = "; cin>>m; } while (n<1 || n>max || m<1 || m>max); cout<<"Nhap cac gia tri cua mang: "<<endl; for (i = 0; i<n; i++) for (j = 0; j<m; j++) { cin>>a[i][j]; } cout<<"Cac PT cua mang la: "<<" "; for (i = 0; i<n; i++) { cout<<endl; for (j = 0; j<m; j++) cout<<a[i][j]<<" "; } cout<<"\n Cac PT chan cua mang la: "; for (i = 0; i<n; i++) for (j = 0; j<m; j++) {if (a[i][j] % 2 == 0) cout<<a[i][j]<<"\t";} cout<<"\n Cac PT le cua mang la: "; for (i = 0; i<n; i++) for (j = 0; j<m; j++) {if (a[i][j] % 2 != 0) cout<<a[i][j]<<"\t";} } Bi tp luyn Bi 1: Vit chng trnh nhp vo mng N s nguyn (N<100) + In cc phn t trong mng + m v in s phn t chn trong mng + m v in s phn t l trong mng Bi 2: Vit chng trnh nhp vo mng N s nguyn (N<100) + Tnh tng v in tt c cc phn t chn trong mng + Tnh tng v in tt c cc phn t l trong mng + Tnh tng v in tt c cc phn t m trong mng + Tnh tng v in tt c cc phn t l khng m trong mng Bi 3:Vit chng trnh nhp vo mng N s nguyn (N<100) + Kim tra mng c sp th t tng khng ? + Kim tra mng c sp th t gim khng ? + Sp xp mng theo th t tng (nu mng cha c sp tng) + Sp xp mng theo th t gim (nu mng cha c sp th t gim) Chng 7. Con tr 1. Khi nim Con tr l mt bin dng  cha a ch. V c nhiu loi a ch nn cing c nhiu kiu con tr tng ng. Kiu con tr int dng  cha a ch bin kiu int. Tng t ta c con tr kiu float, kiu double,& Cing nh vi 1 bin bt k no khc, con tr cn c khai bo trc khi s dng. 2. Ton t ly a ch (&) a ch: Khi khai bo bin, my s cp pht cho bin mt khong nh. a ch ca bin l s th t ca byte u tin trong mt dy cc byte lin tip m my dnh cho bin (cc byte c nh s t 0). My phn bit cc loi a ch nh cc bin. V d nh: a ch kiu int, kiu float,& Vo thi im m chng ta khai bo mt bin th n phi c lu tr trong mt v tr c th trong b nh. Chng ta khng quyt nh ni no bin  c t, iu  lm t ng bi trnh bin dch v h iu hnh. Nhng khi h iu hnh  gn mt a ch cho bin th chng ta cn bit bin  c lu tr  u. iu ny c thc hin bng cch t trc tn bin mt du v (&). V d: x = &y; Gii thch: Gn cho bin x a ch ca bin y v khi t trc tn bin y du & ta khng ni n ni dung ca bin na m ch ni n a ch ca n trong b nh. Gi s bin y c t trng nh c a ch l 1202 v c cc cu lnh nh sau: y = 30; z = y; x = &y; Kt qu: z = 30; x = 1202; Ch : C th nh ngh)a con tr nh sau: Nhng bin lu tr a ch ca bin khc c gi l con tr.  v d trn, bin x l con tr. 3. Ton t tham chiu (*) Bng cch s dng con tr chng ta c th truy xut trc tip n gi tr c lu tr trong bin c tr bi n bng cch t trc tn bin con tr mt du (*). V d: Gi s bin y c t trong nh c a ch l 1202 v c cc cu lnh nh sau: y = 30; z = y; x = &y; t = *y; Kt qu: bin t s mang gi tr l 30. 4. Khai bo bin kiu con tr. V con tr c kh nng tham chiu trc tip n gi tr m chng tr ti nn cn thit phi ch r kiu d liu no m mt bin con tr tr ti khi khai bo. Cu trc khai bo: Kiu_d_liu *Tn_con_tr; Ch : kiu d liu  y l kiu d liu c tr ti, khng phi l kiu ca bn thn con tr. V d: int *x; //Khai bo con tr x kiu int float *t; //Khai bo con tr t kiu float Ch : du (*) khi khai bo bin kiu con tr ch c ngh)a rng  l mt con tr, khng lin quan n ton t tham chiu. V d n gin v s dng con tr: khai bo, s dng ton t tham chiu, ton t ly a ch: #include<iostream.h> main() { int x1 = 5, x2 = 15; int *y; y = &x1; *y = 10; y = &x2; *y = 20; cout<< Gia tri 1 =  <<x1<< \n Gia tri 2 =  <<x2; return 0; } Kt qu: x1 = 10; x2 = 20; 5. Cc php ton C bn nhm php ton lin quan n con tr v a ch l: php gn, php tng gim a ch, php truy nhp b nh v php so snh. Kiu con tr v kiu a ch c vai tr quan trng trong cc php ton ni trn. 5.1. Php gn php gn ch nn thc hin php gn cho con tr cng kiu. Mun gn cc con tr khc kiu phi dng php p kiu nh v d sau: int x; char *pc; pc = (char*)(&x); //p kiu 5.2. Php tng gim a ch //Gii thch php tng gim a ch qua cc v d. V d 1: Cc cu lnh: float x[30], *px; px = &x[10]; Cho bit px l con tr float tr ti phn t x[10]. Kiu a ch float l kiu a ch 4 byte (mi gi tr float cha trong 4 byte), nn cc php tng gim a ch c thc hin trn 4 byte. Ni cch khc: px + i tr ti phn t x[10+i]; px  i tr ti phn t x[10-i]; V d 2: float b[40][50]; b tr ti u dng th nht (phn t b[0][0]; b + 1 tr ti u dng th hai (phn t b[1][0]; & . 5.3. Php truy nhp b nh C nguyn tc l con tr float truy nhp ti 4 byte, con tr int truy nhp 2 byte, chon tr char truy nhp 1 byte. V d: float *pf; int *pi; char *pc; Khi : Nu pf tr n byte th 10001, th *pf biu th vng nh 4 byte lin tip t byte 10001 n 10004. Nu pi tr n byte th 10001, th *pi biu th vng nh 2 byte l 10001 v 10002. Nu pc tr n byte th 10001, th *pc biu th vng nh 1 byte l byte 10001. 5.4. Php so snh Cho php so snh cc con tr cng kiu, v d nu p1 v p2 l 2 con tr float th: p1<p2 nu a ch p1 tr ti thp hn a ch p2 tr ti. p1=p2 nu a ch p1 tr ti bng a ch p2 tr ti. p1<p2 nu a ch p1 tr ti cao hn a ch p2 tr ti. 6. Con tr hng V d c khai bo sau: int a[20]; int *p; Lnh sau s hp l: p = a;  y p v a l tng ng v chng c cng thuc tnh, s khc bit duy nht l chng ta c th gn mt gi tr khc cho con tr p trong khi a lun tr n phn t u tin trong s 20 phn t kiu int m n c nh ngh)a. V vy khng ging nh p   l mt bin con tr bnh thng, a, l mt con tr hng. Lnh gn sau y l khng ng: a = p; bi v a l mt mng (con tr hng) v khng c gi tr no c th c gn cho cc hng. 7. Con tr mng Trong thc t, tn ca mt mng tng ng vi a ch phn t u tin ca n, ging nh mt con tr tng ng vi a ch ca phn t u tin m n tr ti, v vy thc t chng hon ton nh nhau. V d, cho hai khai bo sau: int numbers[20]; int * p; Lnh sau s hp l: p = numbers;  y p v numbers l tng ng v chng c cing thuc tnh, s khc bit duy nht l chng ta c th gn mt gi tr khc cho con tr p trong khi numbers lun tr n phn t u tin trong s 20 phn t kiu int m n c nh ngh)a vi. V vy, khng ging nh p -  l mt bin con tr bnh thng, numbers l mt con tr hng. Lnh gn sau y l khng hp l: numbers = p; bi v numbers l mt mng (con tr hng) v khng c gi tr no c th c gn cho cc hng. V con tr cing c mi tnh cht ca mt bin nn tt c cc biu thc c con tr trong v d di y l hon ton hp l: #include <iostream.h> int main () { int a[5]; int * p; p = a; *p = 10; p++; *p = 20; p = &a[2]; *p = 30; p = a + 3; *p = 40; p = a; *(p+4) = 50; for (int n=0; n<5; n++) cout << a[n] << ", "; return 0; } Bi tp luyn tp: Bi 1: Tnh tng cc phn t ca mng 1 chiu. Cc phn t c nhp t bn phm. Bi 2: 8. Khi to con tr Khi khai bo con tr c th chng ta s mun ch nh r rng chng s tr ti bin no: int number; int *tommy = &number; l tng ng vi: int number; int *tommy; tommy = &number; Trong mt php gn con tr chng ta phi lun lun gn a ch m n tr ti ch khng phi l gi tr m n tr ti. Cn phi nh rng khi khai bo mt bin con tr, du sao (*) c dng  ch ra n l mt con tr, v hon ton khc vi ton t tham chiu.  l hai ton t khc nhau mc d chng c vit vi cng mt du. V vy, cc cu lnh sau l khng hp l: int number; int *tommy; *tommy = &number; Nh i vi mng, trnh bin dch cho php chng ta khi to gi tr m con tr tr ti bng gi tr hng vo thi im khai bo bin con tr: char * terry = "hello"; trong trng hp ny mt khi nh t)nh c dnh  cha "hello" v mt con tr tr ti k t u tin ca khi nh ny ( l k t h') c gn cho terry. Nu "hello" c lu ti a ch 1702, lnh khai bo trn c th c hnh dung nh th ny:  INCLUDEPICTURE "http://www.lamdong.gov.vn/truongkt/thuvien/TailieuvaC++hung/HINH/imgpoin3.gif" \* MERGEFORMATINET  Cn phi nhc li rng terry mang gi tr 1702 ch khng phi l 'h' hay "hello". Bin con tr terry tr ti mt xu k t v n c th c s dng nh l i vi mt mng (hy nh rng mt mng ch n thun l mt con tr hng). V d, nu chng ta mun thay k t 'o' bng mt du chm than, chng ta c th thc hin vic  bng hai cch: terry[4] =  ! ; *(terry+4) = '!'; Hy nh rng vit terry[4] l hon ton ging vi vit *(terry+4) mc d biu thc thng dng nht l ci u tin. Vi mt trong hai lnh trn xu do terry tr n s c gi tr nh sau:  INCLUDEPICTURE "http://www.lamdong.gov.vn/truongkt/thuvien/TailieuvaC++hung/HINH/imgpoin4.gif" \* MERGEFORMATINET  9. Con tr tr ti con tr C++ cho php s dng cc con tr tr ti cc con tr khc ging nh l tr ti d liu.  lm vic  chng ta ch cn thm mt du sao (*) cho mi mc tham chiu. char a; char *b; char **c; a =  z ; b = &a; c = &b; gi s rng a,b,c c lu  cc nh 7230, 8092 and 10502, ta c th m t on m trn nh sau:  INCLUDEPICTURE "http://www.lamdong.gov.vn/truongkt/thuvien/TailieuvaC++hung/HINH/imgpoin6.gif" \* MERGEFORMATINET  im mi trong v d ny l bin c, chng ta c th ni v n theo 3 cch khc nhau, mi cch s tng ng vi mt gi tr khc nhau: c l mt bin c kiu char ** mang gi tr 8092; *c l mt bin c kiu char * mang gi tr l 7230; **c l mt bin c kiu char mang gi tr  z ; 10. Con tr khng kiu Con tr khng kiu l mt loi con tr c bit. N c th tr ti bt k loi d liu no, t gi tr nguyn hoc thc cho ti mt xu k t. Hn ch duy nht ca n l d liu c tr ti khng th c tham chiu ti mt cch trc tip (chng ta khng th dng ton t tham chiu * vi chng) v  di ca n l khng xc nh v v vy chng ta phi dng n ton t chuyn kiu d liu hay php gn  chuyn con tr khng kiu thnh mt con tr tr ti mt loi d liu c th. Mt trong nhng tin ch ca n l cho php truyn tham s cho hm m khng cn ch r kiu. V d: #include <iostream.h> void increase (void* data, int type) { switch (type) { case sizeof(char) : (*((char*)data))++; break; case sizeof(short): (*((short*)data))++; break; case sizeof(long) : (*((long*)data))++; break; } } int main () { char a = 5; short b = 9; long c = 12; increase (&a,sizeof(a)); increase (&b,sizeof(b)); increase (&c,sizeof(c)); cout << (int) a << ", " << b << ", " << c; return 0; } Kt qu: 6, 10, 13 sizeof l mt ton t ca ngn ng C++, n tr v mt gi tr hng l kch thc tnh bng byte ca tham s truyn cho n, v d sizeof(char) bng 1 v kch thc ca char l 1 byte. 11. Con tr hm C++ cho php thao tc vi cc con tr hm. Tin ch tuyt vi ny cho php truyn mt hm nh l mt tham s n mt hm khc.  c th khai bo mt con tr tr ti mt hm chng ta phi khai bo n nh l khai bo mu ca mt hm nhng phi bao trong mt cp ngoc n () tn ca hm v chn du sao (*) ng trc. V d: #include <iostream.h> int addition (int a, int b) { return (a+b); } int subtraction (int a, int b) { return (a-b); } int (*minus)(int,int) = subtraction; int operation (int x, int y, int (*functocall)(int,int)) { int g; g = (*functocall)(x,y); return (g); } int main () { int m,n; m = operation (7, 5, &addition); n = operation (20, m, minus); cout <<n; return 0; } Trong v d ny, minus l mt con tr ton cc tr ti mt hm c hai tham s kiu int, con tr ny c gm  tr ti hm subtraction, tt c u trn mt dng: int (* minus)(int,int) = subtraction; Chng 8. Cu trc 1. Khi nim cu trc Cu trc (struct) thc cht l mt kiu d liu do ngi dng nh ngh)a bng cch gom nhm cc kiu d liu c bn c sn trong C thnh mt kiu d liu phc hp nhiu thnh phn. Cu trc gip cho vic t chc cc d liu phc tp, c bit trong nhng chng trnh ln v trong nhiu tnh hung chng cho php nhm cc bin c lin quan li  x l nh mt n v thay v cc thc th tch bit. Mt v d l cu trc Sinh vin, trong  mi sinh vin c m t bi tp hp cc thuc tnh nh: H tn, Ngy sinh, a ch, Tn lp, im,.. mt s trong cc thuc tnh ny li c th l cu trc bi trong n c th cha nhiu thnh phn nh: H tn gm: h, tn m, tn; a ch: x (phng, th trn), huyn (qun), tnh (thnh ph)& 2. Khai bo cu trc 2.1. Kiu cu trc Khi xy dng cu trc, ta cn m t kiu ca n. iu ny cing tng t nh vic phi thit k ra mt kiu nh trc khi i xy dng cn nh thc s. Cng vic nh ngh)a mt kiu cu trc bao gm vic nu ra tn ca kiu cu trc v cc thnh phn ca n theo 2 cch sau: Cch 1: struct tn_kiu_cu_trc { Khai bo cc thnh phn ca cu trc }; Trong : struct: l t kha. Tn_kiu_cu_trc: l mt tn bt k do ngi lp trnh t t theo quy nh t tn ca C/C++. Thnh phn ca cu trc c th l: bin, mng, cu trc khc  c nh ngh)a trc ,& V d 1: struct Sinhvien { char Hoten; float Diem; char Diachi; }; V d trn nh ngh)a kiu cu trc c tn l Sinhvien bao gm 3 thnh phn: Hoten c kiu char, Diem c kiu float v Diachi c kiu char. V d 2: struct Ngay { int ngaythu; char thang; int nam; }; nh ngh)a cu trc lng nhau: struct SV { char Hoten; struct Ngay Ngaysinh; struct Ngaynhaphoc; float Diem; }; Cch 2: nh ngh)a cu trc bng typedef C th dng t kha typedef  nh ngh)a kiu cu trc ca cc v d  trn nh sau: V d 1: typedef struct { char Hoten; float Diem; char Diachi; } Sinhvien; V d 2: typedef struct { int ngaythu; char thang; int nam; } Ngay; nh ngh)a cu trc lng nhau: typedef struct { char Hoten; struct Ngay Ngaysinh; struct Ngaynhaphoc; float Diem; }SV; 2.2. Khai bo thnh phn (bin, mng) kiu cu trc Khi ta nh ngh)a kiu d liu tc l ta c mt kiu d liu mi, mun s dng ta phi khai bo bin hoc mng. C php khai bo kiu d liu cing ging nh cch khai bo ca cc kiu d liu chun. Cu trc khai bo: + Nu  nh ngh)a cu trc trc, c th khai bo bin theo c php sau: struct < tn cu trc > < tn bin > ; V d: Theo cu trc  nh ngh)a  trn ta c th khai bo bin nh sau: struct Sinhvien SV1, SV2; V d trn khai bo 2 bin kiu Sinhvien c tn l: SV1, SV2. + Hoc c th nh ngh)a bin kiu cu trc ngay khi to cu trc mi: V d: struct Sinhvien { char Hoten; struct Ngay Ngaysinh; struct Ngaynhaphoc; float Diem; }SV1, Sv2; 3. Truy cp n cc thnh phn ca cu trc  truy cp n mt thnh phn (l bin hoc mng) ca mt cu trc ta s dng mt trong cc cch vit sau: Tn cu trc.tn thnh phn; Tn cu trc.tn cu trc.tn thnh phn; Tn cu trc.Tn cu trc.tn cu trc.tn thnh phn; & . Cch vit th nht c s dng khi bin hoc mng l thnh phn trc tip ca mt cu trc. Cc cch vit cn li c s dng khi bin hoc mng l thnh phn trc tip ca mt cu trc m bn thn cu trc ny li l thnh phn ca cu trc ln hn. V d: cout<<SV1.Hoten; // s in h tn ca sinh vin 1 ln mn hnh. cout<<SV2.Ngaynhaphoc.nam; 4. V d cu trc V d 1: Gi s mi gio vin c cc thnh phn: + H tn + Nm sinh + Nnm vo ngnh + mc lng + C gia nh hay cha? Yu cu vit mt chng trnh : + Xy dng cu trc c s d liu cho cc gio vin vn v ton. + Nhp d liu cho 2 gio vin . + Xut d liu ca 2 gio vin  ra mn hnh. & & & & & & & #include<iostream.h> char hoi; main() { struct giaovien { char hoten[40]; int namsinh; int namvaonganh; float mucluong; char cogiadinh; }; giaovien gvtoan; giaovien gvvan; cout<<"Nhap ho ten giao vien toan (khong nhap khoang trang): "; cin>>gvtoan.hoten; cout<<"Nam sinh: "; cin>>gvtoan.namsinh; cout<<"Nam vao nganh: "; cin>>gvtoan.namvaonganh; cout<<"Muc luong: "; cin>>gvtoan.mucluong; cout<<"co gia dinh c/k: "; cin>>hoi; if (hoi == 'c') gvtoan.cogiadinh = 'c'; else gvtoan.cogiadinh = 'k'; cout<<"Nhap ho ten giao vien van (khong nhap khoang trang): "; cin>>gvvan.hoten; cout<<"Nam sinh: "; cin>>gvvan.namsinh; cout<<"Nam vao nganh: "; cin>>gvvan.namvaonganh; cout<<"Muc luong: "; cin>>gvvan.mucluong; cout<<"co gia dinh c/k: "; cin>>hoi; if (hoi == 'c') gvvan.cogiadinh = 'c'; else gvvan.cogiadinh = 'k'; // Xuat ra mn hnh cout<<"\n Danh sch gio vin"; cout<<"\n Ho tn gv ton: "<<gvtoan.hoten; cout<<"\n Nam sinh: "<<gvtoan.namsinh; cout<<"\n Nam vao nganh: "<<gvtoan.namvaonganh; cout<<"\n Muc luong: "<<gvtoan.mucluong; cout<<"\n Co gia dinh: "<<gvtoan.cogiadinh; cout<<"\n __________________________"; cout<<"\n Ho tn giao vien Van: "<<gvvan.hoten; cout<<"\n Nam sinh: "<<gvvan.namsinh; cout<<"\n Nam vao nganh: "<<gvvan.namvaonganh; cout<<"\n Muc luong: "<<gvvan.mucluong; cout<<"\n Co gia dinh: "<<gvvan.cogiadinh; return 0; } V d 2: Lp trnh qun l thng tin cn b. Bit rng mi cn b c cc thnh phn nh sau: + H tn; + Ngy thng nm sinh; + Ngy vo c quan; + Bc lng. Yu cu: vit mt chng trnh : + Vo s liu ca mt cn b; + a s liu  ln mn hnh. & & & ... #include<iostream.h> typedef struct { int ngay; char thang[10]; int nam; } date; typedef struct { char hoten[30]; date ngaysinh; date ngayvaocq; float luong; } canbo; main() { canbo p; cout<<"\n Ho va ten: "; cin>>p.hoten; cout<<"\n Sinh ngay : "; cin>>p.ngaysinh.ngay; cout<<"\n Thang : "; cin>>p.ngaysinh.thang; cout<<"\n Nam : "; cin>>p.ngaysinh.nam; cout<<"\n Vao co quan ngay : "; cin>>p.ngayvaocq.ngay; cout<<"\n Thang : "; cin>>p.ngayvaocq.thang; cout<<"\n Nam : "; cin>>p.ngayvaocq.nam; cout<<"\n Luong : "; cin>>p.luong; cout<<"\n Ngay sinh:"<<p.ngaysinh.ngay<<"/"<<p.ngaysinh.thang<<"/"<<p.ngaysinh.nam; cout<<"\n Ngay vao co quan:"<<p.ngayvaocq.ngay<<"/"<<p.ngayvaocq.thang<<"/"<<p.ngayvaocq.nam; cout<<"\n Luong : "<<p.luong; } Bi tp luyn: Bi 1: T chc d liu  qun l sinh vin bng cu trc mu tin trong mt mng N phn t, mi phn t c cu trc nh sau: - M sinh vin. - Tn. - Nm sinh. - im ton, l, ho, im trung bnh. Vit chng trnh thc hin nhng cng vic sau: " Nhp danh sch cc sinh vin cho mt lp hc. " Xut danh sch sinh vin ra mn hnh. " Tm sinh vin c im trung bnh cao nht. " Sp xp danh sch lp theo th t tng dn ca im trung bnh. " Sp xp danh sch lp theo th t gim dn ca im ton. Tm kim v in ra cc sinh vin c im trung bnh ln hn 5 v khng c mn no di 3. " Tm sinh vin c tui ln nht. " Nhp vo tn ca mt sinh vin. Tm v in ra cc thng tin lin quan n sinh vin  (nu c). Bi 2: T chc d liu qun l danh mc cc b phim VIDEO, cc thng tin lin quan n b phim ny nh sau: - Tn phim (ta phim). - Th loi (3 loi : hnh s, tnh cm, hi). - Tn o din. - Tn in vin nam chnh. - Tn din vin n chnh. - Nm sn xut. - Hng sn xut Vit chng trnh thc hin nhng cng vic sau : " Nhp vo b phim mi cng vi cc thng tin lin quan n b phim ny. " Nhp mt th loi: In ra danh sch cc b phim thuc th loi ny. " Nhp mt tn nam din vin. In ra cc b phim c din vin ny ng. " Nhp tn o din. In ra danh sch cc b phim do o din ny dn dng. Bi 3: Mt th vin cn qun l thng tin v cc u sch. Mi u sch bao gm cc thng tin sau : MaSSach (m s sch), TenSach (tn sch), TacGia (tc gi), SL (s lng cc cun sch ca u sch). Vit chng trnh thc hin cc chc nng sau: " Nhp vo mt danh sch cc u sch (ti a l 100 u sch). " Nhp vo tn ca quyn sch. In ra thng tin y  v cc sch c tn , nu khng c th tn ca quyn sch  th bo l :Khng Tm Thy. " Tnh tng s sch c trong th vin. Chng 9. File 1. Khi nim File Tp tin hay tp d liu l mt tp hp cc d liu c lin quan vi nhau v c cng mt kiu c nhm li thnh mt dy. Chng thng c cha trong mt thit b nh ngoi ca my tnh ()a mm, )a cng,& ) di mt ci tn no . Tp l mt kiu d liu c cu trc. nh ngh)a ca tp c phn no ging mng  ch chng u l tp hp ca cc phn t d liu c cng kiu. Song mng c nh ngh)a v khai bo trong chng trnh vi s phn t  c xc nh cn s phn t ca tp khng c xc nh khi nh ngh)a. Tp c cha trong b nh ngoi, iu  c ngh)a l tp c lu tr  dng nhiu ln v tn ti ngay c khi chng trnh kt thc hoc mt in. C 2 loi tp tin: + Tp tin vn bn: l tp tin dng  ghi cc k t ln )a theo cc dng. Tp tin vn bn l tp cha cc phn t k t l cc ch ci (vit hoa, vit thng); ch s: ;0 ,  1 , & ,  9 ; cc du chm cu, khng k cc k t iu khin. Cc k t ny c t chc thnh tng dng vi du kt thc dng l CR ( Carriage Return - V u dng, m Ascii l 13 )=  \r v LF (Line Feed - Xung dng, m Ascii l 10) =  \n , tp vn bn dng k t ^Z (xc nh bi t hp phm Ctrl_Z) c m ASCII l 26  lm k hiu kt thc tp. V vy, tp vn bn c th c c trn mn hnh, c th son tho vi cc phn mm son tho vn bn. + Tp tin nh phn: l tp tin dng  ghi cc cu trc dng nh phn (c m ha). Tp tin nh phn cha kh nhiu d liu c m iu khin l cc k t c m ASCII t 0 n 31. Nu mt xu k t c ghi ra tp nh phn th k hiu ht dng ca n ch cn c LF =  \n , khng c k t CR =  \r . 2. To file c file Qu trnh thao tc trn tp tin thng qua 4 bc: Bc 1: Khai bo con tr tr n tp tin. Bc 2: M tp tin. Bc 3: Cc x l trn tp tin. Bc 4: ng tp tin. 2.1. Khai bo con tr tr n tp tin C php: FILE *Tn_bin; V d: FILE *f; //Khai bo bin con tr file f; FILE *vb,*np; //Khai bo 2 bin con tr tp. Ch : + Trc khi khai bo con tr tr n tp tin, cn khai bo th vin: stdio.h. + Mi tp u c kt thc bng mt du hiu c bit  bo hiu ht tp, hay gi l EOF (End of File). Gi tr EOF trong UNIX c nh ngh)a l -1, trong chng trnh dch C/C++ khc c th c cc gi tr khc. + C/C++ c mt hm chun feof, theo kiu Boolean, vi tham s l mt bin tp  th xem ca s  t vo v tr kt thc tp  cha. Nu ca s cn cha tr vo phn t cui tp th feof c gi tr l False, tc l bng 0. S dng hm feof() s an ton v chnh xc hn. 2.2. M tp tin Cu trc: Bin_tp = fopen (<ng dn ca tp tin>,<kiu truy nhp>); V d 1: FILE *f; //Khai bo bin con tr f f = fopen ( C:\\VD1.txt ,  rt ); V d 2: M mt tp tin tn TEST.txt  ghi. FILE *f; f = fopen( TEST.txt ,  w ); if (f!=NULL) { /* Cc cu lnh  thao tc vi tp tin*/ /* ng tp tin*/ } Trong v d trn, ta c s dng cu lnh kim tra iu kin  xc nh m tp tin c thnh cng hay khng?. Nu m tp tin  ghi, nu tp tin  tn ti ri th tp tin s b xa v mt tp tin mi c to ra. Nu ta mun ghi ni d liu, ta phi s dng ch   a . Khi m vi ch  c, tp tin phi tn ti ri, nu khng mt li s xut hin. 2.3. Cc kiu x l tp thng dng + Cho tp vn bn: STTKiu ngh)ar, rtM mt tp  c theo kiu vn bn. Tp cn phi tn ti.w, wtM mt tp  ghi theo kiu vn bn. Nu tp  tn ti th n s b xa.a, atM mt tp  ghi b sung theo kiu vn bn. Nu tp cha tn ti th to tp mi.r+M mt tp  c theo kiu vn bn. Tp phi tn ti.w+M mt tp  ghi theo kiu vn bn. Tp phi tn ti.a+M mt tp  c/ghi b sung theo kiu vn bn. Nu tp cha tn ti th to tp mi. + Cho tp nh phn: STTKiu ngh)arbM mt tp  c theo kiu nh phn. Tp cn phi tn ti.wbM mt tp mi  ghi theo kiu nh phn. Nu tp  tn ti th n s b xa.abM mt tp  ghi b sung theo kiu nh phn. Nu tp cha tn ti th to tp mi. r+ bM mt tp  c theo kiu vn bn. Tp phi tn ti.w+ bM mt tp  ghi theo kiu vn bn. Tp phi tn ti.a+ bM mt tp  c/ghi b sung theo kiu vn bn. Nu tp cha tn ti th to tp mi.2.4. ng tp tin Hm fclose() c dng  ng tp tin c m bi hm fopen(). Hm ny s ghi d liu cn li trong vng m vo tp tin v ng li tp tin. C php: int fclose(FILE *f) Trong  f l con tr tp tin c m bi hm fopen(). Gi tr tr v ca hm l 0 bo rng vic ng tp tin thnh cng. Hm tr v EOF nu c xut hin li. Ngoi ra, ta cn c th s dng hm fcloseall()  ng tt c cc tp tin li. C php: int fcloseall() Kt qu tr v ca hm l tng s cc tp tin c ng li. Nu khng thnh cng, kt qu tr v l EOF. 2.5. Kim tra n cui tp tin hay cha C php: int feof(FILE *f) ngh)a: Kim tra xem  chm ti cui tp tin hay cha v tr v EOF nu cui tp tin c chm ti, ngc li tr v 0. 2.6. Cc x l trn tp tin + Cc hm  c tp tin vn bn STTTn hm ngh)a1fscanf(FILE *tn_CT,cc tham bin)c d liu t mt tp tin.2fgets(vng nh,kch thc ti a, FILE *tn_CT)c mt chui k t t mt tp tin vi kch thc ti a cho php hoc gp k t xung dng.3getc(FILE *tn_CT)c mt k t t tp tin ang m. + Cc hm  ghi tp tin vn bn STTTn hm ngh)a1fprintf(FILE *tn_CT,cc tham bin)Ghi d liu vo tp tin.2fputs(chui k t, FILE *tn_CT)Ghi mt chui k t vo tp tin ang m. + Cc hm  c tp tin nh phn STTTn hm ngh)a1fread(ptr,sizeof,len, FILE *tn_CT)ptr: vng nh  lu tr d liu c sizeof: kch thc mi nh, tnh bng byte. len:  di d liu cn c. FILE: c t tp tin nh phn no + Cc hm  ghi tp tin nh phn STTTn hm ngh)a1fwirte(&ptr,sizeof,len, FILE *tn_CT)ptr: vng nh  lu tr d liu c sizeof: kch thc mi nh, tnh bng byte. len:  di d liu cn c. FILE: c t tp tin nh phn no  cc phn tip theo, s  cp k hn n cc hm trn qua v d. 2.7. Truy cp n tp tin vn bn (text) Ghi d liu ln tp tin vn bn + Hm putc() Hm ny c dng  ghi mt k t ln mt tp tin vn bn ang c m  lm vic. C php: int putc(int c, FILE *f) Trong , tham s c cha m Ascii ca mt k t no . M ny c ghi ln tp tin lin kt vi con tr f. Hm ny tr v EOF nu gp li. + Hm fputs() Hm ny dng  ghi mt chui k t cha trong vng m ln tp tin vn bn. C php:int puts(const char *buffer, FILE *f) Trong , buffer l con tr c kiu char ch n v tr u tin ca chui k t c ghi vo. Hm ny tr v gi tr 0 nu buffer cha chui rng v tr v EOF nu gp li. + Hm fprintf() Hm ny dng  ghi d liu c nh dng ln tp tin vn bn. C php: fprintf(FILE *f, const char *format, varexpr) Trong :format: chui nh dng, varexpr: danh sch cc biu thc, mi biu thc cch nhau du phy (,). V d: Vit chng trnh ghi chui k t ln tp tin vn bn D:\\Baihat.txt #include<stdio.h> #include<conio.h> int main() { FILE *f; clrscr(); f=fopen("D:\\Baihat.txt","r+"); if (f!=NULL) { fputs("Em oi Ha Noi pho.\n",f); fputs("Ta con em, mui hoang lan; ta con em, mui hoa sua.",f); fclose(f); } getch(); return 0; } Ni dung tp tin Baihat.txt khi c m bng trnh son tho vn bn Notepad:  c d liu t tp tin vn bn + Hm getc() Hm ny dng  c d liu t tp tin vn bn ang c m  lm vic. C php: int getc(FILE *f) Hm ny tr v m Ascii ca mt k t no  (k c EOF) trong tp tin lin kt vi con tr f. + Hm fgets() C php: char *fgets(char *buffer, int n, FILE *f) Hm ny c dng  c mt chui k t t tp tin vn bn ang c m ra v lin kt vi con tr f cho n khi c  n k t hoc gp k t xung dng  \n (k t ny cing c a vo chui kt qu) hay gp k t kt thc EOF (k t ny khng c a vo chui kt qu). Trong : - buffer (vng m): con tr c kiu char ch n cng nh  ln cha cc k t nhn c. - n: gi tr nguyn ch  di ln nht ca chui k t nhn c. - f: con tr lin kt vi mt tp tin no . - K t NULL ( \0 ) t ng c thm vo cui chui kt qu lu trong vng m. - Hm tr v a ch u tin ca vng m khi khng gp li v cha gp k t kt thc EOF. Ngc li, hm tr v gi tr NULL. + Hm fscanf() Hm ny dng  c d liu t tp tin vn bn vo danh sch cc bin theo nh dng. C php:fscanf(FILE *f, const char *format, varlist) Trong : format: chui nh dng (ging hm scanf()); varlist: danh sch cc bin mi bin cch nhau du phy (,). V d: Vit chng trnh chp tp tin D:\Baihat.txt  trn sang tp tin D:\Baica.txt. #include<stdio.h> #include<conio.h> int main() { FILE *f1,*f2; clrscr(); f1=fopen("D:\\Baihat.txt","rt"); f2=fopen("D:\\Baica.txt","wt"); if (f1!=NULL && f2!=NULL) { int ch=fgetc(f1); while (! feof(f1)) { fputc(ch,f2); ch=fgetc(f1); } fcloseall(); } getch(); return 0; } Ni dung tp tin Baica.txt khi c m bng trnh son tho vn bn Notepad:  Ch : Hai tp  Baihat.txt v  Baica.txt u  c trn  D: 3. To file nh phn Sau khi m tp mi xong vi ch  x l l w (writing), tp s rng v cha c phn t no, ca s tp s khng c gi tr xc nh v n tr vo cui tp (EOF). ng thi cn lu khi m tp nu trn b nh ngoi m  c sn tp c tn trng vi tn tp c m th tp ci s b xa i. V d: To ra tp tin c 100 phn t c gi tr t 1 n 100: #include<stdio.h> main() { char filename[21]; int i; long num; FILE *f; printf ( Tn tp tin cn to ra l:  ); scanf ( %20s , filename); f = fopen (filename, wb ); for (i = 1; i<=100; i++) fwrite(&i, sizeof(int),i,f); fclose(f); } 4. c file nh phn 4.1. Ghi d liu ln tp nh phn Ghi d liu ln tp tin nh phn - Hm fwrite() C php:size_t fwrite(const void *ptr, size_t size, size_t n, FILE *f) Trong : - ptr: con tr ch n vng nh cha thng tin cn ghi ln tp tin. - n: s phn t s ghi ln tp tin. - size: kch thc ca mi phn t. - f: con tr tp tin  c m. - Gi tr tr v ca hm ny l s phn t c ghi ln tp tin. Gi tr ny bng n tr khi xut hin li. 4.2. c d liu t tp tin nh phn - Hm fread() C php:size_t fread(const void *ptr, size_t size, size_t n, FILE *f) Trong : - ptr: con tr ch n vng nh s nhn d liu t tp tin. - n: s phn t c c t tp tin. - size: kch thc ca mi phn t. - f: con tr tp tin  c m. - Gi tr tr v ca hm ny l s phn t  c c t tp tin. Gi tr ny bng n hay nh hn n nu  chm n cui tp tin hoc c li xut hin. 4.3. Di chuyn con tr tp tin - Hm fseek() Vic ghi hay c d liu t tp tin s lm cho con tr tp tin dch chuyn mt s byte, y chnh l kch thc ca kiu d liu ca mi phn t ca tp tin. Khi ng tp tin ri m li n, con tr lun  v tr ngay u tp tin. Nhng nu ta s dng kiu m tp tin l  a  ghi ni d liu, con tr tp tin s di chuyn n v tr cui cng ca tp tin ny. Ta cing c th iu khin vic di chuyn con tr tp tin n v tr ch nh bng hm fseek(). C php:int fseek(FILE *f, long offset, int whence) Trong : + f: con tr tp tin ang thao tc. + offset: s byte cn dch chuyn con tr tp tin k t v tr trc . Phn t u tin l v tr 0. + whence: v tr bt u  tnh offset, ta c th chn im xut pht l:  + Kt qu tr v ca hm l 0 nu vic di chuyn thnh cng. Nu khng thnh cng, 1 gi tr khc 0 ( l 1 m li) c tr v. V d: Vit chng trnh ghi ln tp tin  CacSo.Dat 3 gi tr s (thc, nguyn, nguyn di). Sau  c cc s t tp tin va ghi v hin th ln mn hnh. #include<stdio.h> #include<conio.h> int main() { FILE *f; clrscr(); f=fopen("D:\\CacSo.txt","wb"); if (f!=NULL) { double d=3.14; int i=101; long l=54321; fwrite(&d,sizeof(double),1,f); fwrite(&i,sizeof(int),1,f); fwrite(&l,sizeof(long),1,f); /* Doc tu tap tin*/ rewind(f); fread(&d,sizeof(double),1,f); fread(&i,sizeof(int),1,f); fread(&l,sizeof(long),1,f); printf("Cac ket qua la: %f%d%ld",d,i,l); fclose(f); } getch(); return 0; } Bi tp luyn Bi tp 1: Vit chng trnh to ra 2 tp c tn l  LTCB v  LTNC . Trong chng trnh s dng cc hm: + fopen  m tp. + fputc  ghi mt k t ln tp. + fclose  ng tp. Bi tp 2: Hy vit chng trnh theo yu cu sau: + u tin ghi mt dy k t t  K n  Q ln tp c tn l  Baitap2 . + To tp  BT2.sl v ghi ln tp  BT2.sl 10 k t. Bi tp 3: Hy to mt vn bn text gm 3 dng vi ni dung nh sau: Lap trinh co ban Turbo C Pascal Bi tp 4: + Vit chng trnh nhp cc dng k t t bn phm v ghi ln tp c tn l  Baitap4 . + Vit chng trnh c cc dng k t trn tp  Baitap4 v in ra mn hnh. Bi tp 5: Mi sinh vin cn qun l t nht 2 thng tin: m sinh vin v h tn. Vit chng trnh cho php la chn cc chc nng: nhp danh sch sinh vin t bn phm ri ghi ln tp tin SinhVien.dat, c d liu t tp tin SinhVien.dat ri hin th danh sch ln mn hnh, tm kim h tn ca mt sinh vin no  da vo m sinh vin nhp t bn phm. Hng dn: Ta nhn thy rng mi phn t ca tp tin SinhVien.Dat l mt cu trc c 2 trng: m v h tn. Do , ta cn khai bo cu trc ny v s dng cc hm c/ghi tp tin nh phn vi kch thc mi phn t ca tp tin l chnh kch thc cu trc . #include<stdio.h> #include<conio.h> #include<string.h> typedef struct { char Ma[10]; char HoTen[40]; } SinhVien; void WriteFile(char *FileName) { FILE *f; int n,i; SinhVien sv; f=fopen(FileName,"ab"); printf("Nhap bao nhieu sinh vien? ");scanf("%d",&n); fflush(stdin); for(i=1;i<=n;i++) { printf("Sinh vien thu %i\n",i); printf(" - MSSV: ");gets(sv.Ma); printf(" - Ho ten: ");gets(sv.HoTen); fwrite(&sv,sizeof(sv),1,f); fflush(stdin); } fclose(f); printf("Bam phim bat ky de tiep tuc"); getch(); } void ReadFile(char *FileName) { FILE *f; SinhVien sv; f=fopen(FileName,"rb"); printf(" MSSV|Ho va ten\n"); fread(&sv,sizeof(sv),1,f); while (!feof(f)) { printf("%s| %s\n",sv.Ma,sv.HoTen); fread(&sv,sizeof(sv),1,f); } fclose(f); printf("Bam phim bat ky de tiep tuc!!!"); getch(); } void Search(char *FileName) { char MSSV[10]; FILE *f; int Found=0; SinhVien sv; fflush(stdin); printf("Ma so sinh vien can tim: ");gets(MSSV); f=fopen(FileName,"rb"); while (!feof(f) && Found==0) { fread(&sv,sizeof(sv),1,f); if (strcmp(sv.Ma,MSSV)==0) Found=1; } fclose(f); if (Found == 1) printf("Tim thay SV co ma %s. Ho ten la: %s",sv.Ma,sv.HoTen); else printf("Tim khong thay sinh vien co ma %s",MSSV); printf("\nBam phim bat ky de tiep tuc!!!"); getch(); } int main() { int c; for (;;) { clrscr(); printf("1. Nhap DSSV\n"); printf("2. In DSSV\n"); printf("3. Tim kiem\n"); printf("4. Thoat\n"); printf("Ban chon 1, 2, 3, 4: "); scanf("%d",&c); if(c==1) WriteFile("d:\\SinhVien.Dat"); else if (c==2) ReadFile("d:\\SinhVien.Dat"); else if (c==3) Search("d:\\SinhVien.Dat"); else break; } return 0; } Ti liu tham kho Gio trnh lp trnh C v C++ 2. Ngn ng lp trnh C v C++ Bi ging  Bi tp  Li gii mu  Ng Trung Vit - NXB Thng k. 3. Gio trnh ngn ng lp trnh C  Phm Vn t  NXB Th gii 4. Ngn ng lp trnh C++ - Quch Tun Ngc  NXB Thng k. 5. Gio trnh C++ v lp trnh hng i tng  Phm Vn t, L Trng Thng  NXB Hng c. 6. Gio trnh K thut lp trnh C c bn v nng cao  NXB Hng c.      cng mn: Lp trnh C bn PAGE  PAGE 42  Kt thc Thm vo danh sch L ngi cui cng? c tn, tui Bt u Nu tui > 50? 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