& & a a * * ptr p int a ; int *a ; int a ; int a int *a

Similar documents
コマンドラインから受け取った文字列の大文字と小文字を変換するプログラムを作成せよ 入力は 1 バイトの表示文字とし アルファベット文字以外は変換しない 1. #include <stdio.h> 2. #include <ctype.h> /*troupper,islower,isupper,tol

ohp03.dvi

r03.dvi

joho07-1.ppt

r08.dvi

file"a" file"b" fp = fopen("a", "r"); while(fgets(line, BUFSIZ, fp)) {... fclose(fp); fp = fopen("b", "r"); while(fgets(line, BUFSIZ, fp)) {... fclose

ohp08.dvi

新・明解C言語 ポインタ完全攻略

ex12.dvi

C B

1.ppt

/ SCHEDULE /06/07(Tue) / Basic of Programming /06/09(Thu) / Fundamental structures /06/14(Tue) / Memory Management /06/1

A/B (2018/10/19) Ver kurino/2018/soft/soft.html A/B

main

PowerPoint Presentation

WinHPC ppt

II ( ) prog8-1.c s1542h017%./prog8-1 1 => 35 Hiroshi 2 => 23 Koji 3 => 67 Satoshi 4 => 87 Junko 5 => 64 Ichiro 6 => 89 Mari 7 => 73 D

02: 変数と標準入出力

卒 業 研 究 報 告.PDF

lexex.dvi

r07.dvi

ohp07.dvi


gengo1-12

gengo1-12

02: 変数と標準入出力

目 目 用方 用 用 方

‚æ2›ñ C„¾„ê‡Ìš|

£Ã¥×¥í¥°¥é¥ß¥ó¥°ÆþÌç (2018) - Â裵²ó ¨¡ À©¸æ¹½Â¤¡§¾ò·ïʬ´ô ¨¡

I. Backus-Naur BNF : N N 0 N N N N N N 0, 1 BNF N N 0 11 (parse tree) 11 (1) (2) (3) (4) II. 0(0 101)* (

II 3 yacc (2) 2005 : Yacc 0 ~nakai/ipp2 1 C main main 1 NULL NULL for 2 (a) Yacc 2 (b) 2 3 y

double float

17 1. strucr Potter ( ) Harry Potter and the Philosopher s Stone 1997 English Title : Harry Potter and the Philosopher s Stone : : 1997 #include<stdio

Prog1_6th

Original : Hello World! (0x0xbfab85e0) Copy : Hello World! (0x0x804a050) fgets mstrcpy malloc mstrcpy (main ) mstrcpy malloc free fgets stream 1 ( \n

Microsoft Word - no14.docx

para02-2.dvi

ex01.dvi

I. Backus-Naur BNF S + S S * S S x S +, *, x BNF S (parse tree) : * x + x x S * S x + S S S x x (1) * x x * x (2) * + x x x (3) + x * x + x x (4) * *

£Ã¥×¥í¥°¥é¥ß¥ó¥°(2018) - Âè11²ó – ½ÉÂꣲ¤Î²òÀ⡤±é½¬£² –

£Ã¥×¥í¥°¥é¥ß¥ó¥°ÆþÌç (2018) - Â裶²ó ¨¡ À©¸æ¹½Â¤¡§·«¤êÊÖ¤· ¨¡

I J

tuat1.dvi

ex14.dvi


新版明解C言語 実践編

XMPによる並列化実装2

A/B (2018/06/08) Ver kurino/2018/soft/soft.html A/B

comment.dvi

£Ã¥×¥í¥°¥é¥ß¥ó¥°ÆþÌç (2018) - Â裱£²²ó ¡Ý½ÉÂꣲ¤Î²òÀ⡤±é½¬£²¡Ý

橡Pro PDF

2008 ( 13 ) C LAPACK 2008 ( 13 )C LAPACK p. 1

:30 12:00 I. I VI II. III. IV. a d V. VI

< F2D837C E95CF CF68A4A94C5816A2E6A>

Microsoft Word - C.....u.K...doc

Microsoft Word - Cプログラミング演習(12)

:30 12:00 I. I VI II. III. IV. a d V. VI

I117 II I117 PROGRAMMING PRACTICE II 2 SOFTWARE DEVELOPMENT ENV. 2 Research Center for Advanced Computing Infrastructure (RCACI) / Yasuhiro Ohara yasu

超初心者用

BW BW

Microsoft Word - no11.docx

PowerPoint プレゼンテーション



人事行政の運営状況等の公表(平成19年12月)(PDF)


mogiJugyo_slide_full.dvi

(K&R 2.9) ~, &,, >>, << 2. (K&R 5.7) 3. (K&R 5.9) 4. (K&R 5.10) (argc argv atoi(), atof() ) 5. (K&R 7.5) (K&R 7.6) - FILE, stdin, stdout, std

ex01.dvi

Taro-ファイル処理(公開版).jtd

EPSON VP-1200 取扱説明書

プログラミング基礎

‚æ4›ñ

mstrcpy char *mstrcpy(const char *src); mstrcpy malloc (main free ) stdio.h fgets char *fgets(char *s, int size, FILE *stream); s size ( )

PowerPoint プレゼンテーション

yacc.dvi


平成12年度 証券貯蓄に関する全国調査

C言語によるアルゴリズムとデータ構造

P6dark P6dark µ-pic 2 µ-pic 2 3 µ-pic µ-pic µ-pic 3 µ-pic (10cm ) MPGC N3035-KA195 No. SN ASD (16ns[C]) (16nsC

( CUDA CUDA CUDA CUDA ( NVIDIA CUDA I

新たな基礎年金制度の構築に向けて

P06.ppt

program7app.ppt

(search: ) [1] ( ) 2 (linear search) (sequential search) 1

memo

PSCHG000.PS

Condition DAQ condition condition 2 3 XML key value

[1] #include<stdio.h> main() { printf("hello, world."); return 0; } (G1) int long int float ± ±

( ) 1 1: 1 #include <s t d i o. h> 2 #include <GL/ g l u t. h> 3 #include <math. h> 4 #include <s t d l i b. h> 5 #include <time. h>

実際の株価データを用いたオプション料の計算

Microsoft PowerPoint - kougi9.ppt

86 8 MPIBNCpack 15 : int n, myid, numprocs, i; 16 : double pi, start_x, end_x; 17 : double startwtime = 0.0, endwtime; 18 : int namelen; 19 : char pro

「東京こどもネット・ケータイヘルプデスク(こたエール)」平成22年度相談実績の概要

key

5-1_a-kanaoka_JPNICSecSemi_Phish_Tech_ _3.PDF



ポインタ変数

slide5.pptx

Transcription:

int a = 123; a 123 :100 a 123 int *ptr = & a; a ptr ptr a 100 a 123 200 *ptr 200 a

& & a a * * ptr p --------------------------------------------------------------------------------------------- int a ; int *a ; int a ; int a int *a ; a a int --------------------------------------------------------------------------------------------- 20 int *pc ; pc = &a ; int a ; a = 20; 345 *pc a 20

ⅱ int pc int a pc a int a ; int *pc ; pc a pc = &a ; pc a int pc int a pc * *pc a *pc a *pc a 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 #include<stdio.h> int main() { int a,b; int *pc; a = 25; printf(" a :%x(16 \n\n",&a); printf(" b :%x(16 \n\n",&b); printf(" a :%d 10 \n\n",a); puts("-------------------------------------------\n"); pc = &a; b = *pc + 100; *pc = 6; puts("pc = &a\n" "b = *pc +100\n" "*pc =6\n" " \n"); printf(" pc :%x(16 \n\n",pc); printf(" b :%d 10 \n\n",b); printf(" a :%d 10 \n\n",a); return 0;

a :5fbffacc(16 b :5fbffac8 16 a :25 10 ------------------------------------------- pc = &a b = *pc +100 *pc =6 pc b a :5fbffacc 16 :125 10 :6 10 10 11 & 16 12 a 8 25 15 pc a *pc a 21 pc pc a 10 22 b b *pc +100 a+100 125 23 a *pc =a a 8

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 /* */ #include<stdio.h> #include<ctype.h> void replace(char *dest,char *str); int main(int argc,char **argv){ char dest[256]={0; int i; printf(" :%d\n\n",argc); for(i=1; i<argc; i++){ replace(dest, argv[i]); // argv[i] == *(argv + i) printf(" : %s\n : %s\n\n",argv[i],dest); return 0; void replace(char *dest,char *str){ int i; for(i=0; str[i]!= NULL; i++){ if(isalpha(str[i])){ if(isupper(str[i])) dest[i] = tolower(str[i]); else dest[i] = toupper(str[i]); else dest[i] = str[i]; dest[i] = NULL;

Yukihiro-Ikehara-no-MacBook-Pro:prog1 e105746$./a.out Abc DeTrmInaNt. :4 : Abc : abc : DeTrmInaNt : detrminant :. :. i 4 ctype.h isalpha,isupper,tolwer,toupper 6 void replace char *dest, *str 8 19 int main int argc char argv main 2 9 char dest 256 0 12 argc argc 14 17 main for i 1 i argc i +1 16 replace dest argv[i] 17 dest argv[i] 21 40 replace void str for

./a.out Abc DeTrmInaNt. argc argv argv[] argv argv[0]. / a. o u t NULL argv[1] A b c NULL argv[2] D e T r m I n a N t NULL argv[3]. NULL 1 replace argv[0] 2 argv[1]... n argv[n] argv [] NULL 3 replace replace main *dest,*str *str 1 dest NULL main main for i 0./a.out

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 /* */ #include<stdio.h> void reverse(char *dest,char *str); int main(int argc,char **argv){ char dest[256]={0; int i; printf(" :%d\n\n",argc); for(i=1; i<argc; i++){ reverse(dest, argv[i]); printf(" : %s\n : %s\n\n",argv[i],dest); return 0; void reverse(char *dest,char *str){ int i,kiri; for(kiri = 0; str[kiri]!= NULL; kiri++); for(i=0; i<kiri; i++) dest[i]=str[kiri-i-1]; dest[i]=null;

Yukihiro-Ikehara-no-MacBook-Pro:prog1 e105746$./a.out God of eater. :5 : God : dog : of : fo : eater : retae :. :. < main reverse 20 28 14 reverse 20 char *dest *str 14 dest argv[i] for 22 reverse int i,kiri 24 for argv[i] kiri Kiri argv[i] argv[i] God kiri 25 26 argv[i] kiri dest[0] = str[2] dest[1] = str[1] dest[2] = str[0] dest argv[i]

28 dest[i] argv[i] reverse dest str main reverse argv[1] str[0] str[1] str[2] G o d dest[0] dest[1] dest[2] d o G ---------------------------------------------------------- argv[2] str[0] str[1] o f dest[0] dest[1] f o ----------------------------------------------------- argv[3] str[0] str[1] str[2] str[3] str[4] e a t e r

argv[3] dest[0] dest[1] dest[2] dest[3] dest[4] r e t a e ----------------------------------------------------------

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 #include<stdio.h> void reverse(char *dest,char *str); int main(int argc,char **argv){ char dest[256]={0; int i; printf(" :%d\n\n",argc); for(i=1; i<argc; i++){ reverse(dest, argv[i]); printf(" : %s\n : %s\n\n",argv[i],dest); return 0; void reverse(char *dest,char *str){ int i,kiri; for(kiri = 0; str[kiri]!= NULL; kiri++); for(i=0; i<kiri; i++) dest[i]=str[kiri-i-1]; Yukihiro-Ikehara-no-MacBook-Pro:prog1 e105746$./a.out God of eater :4 : God : dog : of : fog : eater : retae reverse dest[i] = NULL; of G

reverse argv[1] str[0] str[1] str[2] G o d dest[0] dest[1] dest[2] d o G ---------------------------------------------------------- argv[2] str[0] str[1] o f G arg1 G f o G arg[3] arg[2] C