strtok

C语言字符串函数大全

时间秒杀一切 提交于 2020-03-31 18:55:35
C语言字符串函数大全 函数名: stpcpy 功 能: 拷贝一个字符串到另一个 用 法 : char *stpcpy(char *destin, char *source); 程序例 : #include <stdio.h> #include <string.h> int main(void) { char string[10]; char *str1 = "abcdefghi"; stpcpy(string, str1); printf("%s\n", string); return 0; } 函数名 : strcat 功 能 : 字符串拼接函数 用 法 : char *strcat(char *destin, char *source); 程序例 : #include <string.h> #include <stdio.h> int main(void) { char destination[25]; char *blank = " ", *c = "C++", *Borland = "Borland"; strcpy(destination, Borland); strcat(destination, blank); strcat(destination, c); printf("%s\n", destination); return 0; } 函数名 : strchr 功

恶心的C语言strtok函数

喜你入骨 提交于 2020-03-29 04:00:05
从C#、JAVA到C ,让我觉得像是从公产主义社会回到了原始社会,不顺手,所以很心里憋气!!! 函数名: strtok 功 能: 查找由在第二个串中指定的分界符分隔开的单词 用 法: char * strtok( char * str1, char * str2); 程序例: #include < string .h > #include < stdio.h > int main( void ) { char input[ 16 ] = " abc,d " ; char * p; /**/ /* strtok places a NULL terminator in front of the token, if found */ p = strtok(input, " , " ); if (p) printf( " %s\n " , p); /**/ /* A second call to strtok using a NULL as the first parameter returns a pointer to the character following the token */ p = strtok(NULL, " , " ); if (p) printf( " %s\n " , p); return 0 ; } 下面是恶心的strtok函数,用作分割字符串

strtok和strtok_r

旧巷老猫 提交于 2020-03-24 07:14:51
strtok和strtok_r 原型:char *strtok(char *s, char *delim); 功能:分解字符串为一组字符串。s为要分解的字符串,delim为分隔符字符串。 说明:首次调用时,s指向要分解的字符串,之后再次调用要把s设成NULL。 strtok在s中查找包括在delim中的字符并用NULL('/0')来替换,直到找遍整个字符串。 返回值 :从s开头開始的一个个被切割的串。当没有被切割的串时则返回NULL。 全部delim中包括的字符都会被滤掉,并将被滤掉的地方设 为一处切割的节点。 举例: #include < string .h > #include < stdio.h > int main( void ) { char input[ 16 ] = " abc,d " ; char * p; /* strtok places a NULL terminator in front of the token, if found */ p = strtok(input, " , " ); if (p) printf( " %s " , p); /* A second call to strtok using a NULL as the first parameter returns a pointer to the character following

strtok和strtok_r

我们两清 提交于 2020-03-21 08:59:18
strtok和strtok_r 原型:char *strtok(char *s, char *delim); 功能:分解字符串为一组字符串。s为要分解的字符串,delim为分隔符字符串。 说明:首次调用时,s指向要分解的字符串,之后再次调用要把s设成NULL。 strtok在s中查找包括在delim中的字符并用NULL('/0')来替换,直到找遍整个字符串。 返回值 :从s开头開始的一个个被切割的串。当没有被切割的串时则返回NULL。 全部delim中包括的字符都会被滤掉,并将被滤掉的地方设 为一处切割的节点。 举例: #include < string .h > #include < stdio.h > int main( void ) { char input[ 16 ] = " abc,d " ; char * p; /* strtok places a NULL terminator in front of the token, if found */ p = strtok(input, " , " ); if (p) printf( " %s " , p); /* A second call to strtok using a NULL as the first parameter returns a pointer to the character following

【C】——strtok()和strtok_r()

时间秒杀一切 提交于 2020-03-18 07:21:19
下面的说明摘自于最新的Linux内核2.6.29,说明了strtok()这个函数已经不再使用,由速度更快的strsep()代替 /* * linux/lib/string.c * * Copyright (C) 1991, 1992 Linus Torvalds */ /* * stupid library routines.. The optimized versions should generally be found * as inline code in <asm-xx/string.h> * * These are buggy as well.. * * * Fri Jun 25 1999, Ingo Oeser <ioe@informatik.tu-chemnitz.de> * - Added strsep() which will replace strtok() soon (because strsep() is * reentrant and should be faster). Use only strsep() in new code, please. * * * Sat Feb 09 2002, Jason Thomas <jason@topic.com.au>, * Matthew Hawkins <matt@mh.dropbear.id.au> * -

字符串函数之Strtok()函数

天大地大妈咪最大 提交于 2020-02-02 03:48:05
Strtok()函数详解:   该函数包含在 "string.h" 头文件中 函数原型: char* strtok (char* str,constchar* delimiters ); 函数功能:   切割字符串,将str切分成一个个子串 函数参数:   str:在第一次被调用的时间str是传入需要被切割字符串的首地址;在后面调用的时间传入NULL。   delimiters:表示切割字符串(字符串中每个字符都会 当作分割符)。 函数返回值:   当s中的字符查找到末尾时,返回NULL;   如果查不到delimiter所标示的字符,则返回当前strtok的字符串的指针。 使用strtok()函数: #include<stdio.h> #include<string.h> int main(void) { char buf[]="hello@boy@this@is@heima"; char*temp = strtok(buf,"@"); while(temp) { printf("%s ",temp); temp = strtok(NULL,"@"); } return0; } 预计输出结果:    "hello boy this is heima " 实际运行结果: 得到的结论:   strtok在切割字符串的时间,实际上就是将分割符的字符delimiter替换为'\0

Reading from CSV with fgets() and strtok()

我与影子孤独终老i 提交于 2020-01-24 13:35:48
问题 FULL DISCLOSURE: SCHOOL ASSIGNMENT I've been working on some code to pull data from a CSV and display it, however I keep encountering this error I can't seem to overcome. The code would be pretty simple, but the problem is that our CSV can have spaces before or after the comma, which means you can't just strtok() because it will include the space. Therefore, you need to remove the spaces before printing. I attempted to achieve this by strtok() the CSV, then running said token through a

Reading from CSV with fgets() and strtok()

自古美人都是妖i 提交于 2020-01-24 13:35:29
问题 FULL DISCLOSURE: SCHOOL ASSIGNMENT I've been working on some code to pull data from a CSV and display it, however I keep encountering this error I can't seem to overcome. The code would be pretty simple, but the problem is that our CSV can have spaces before or after the comma, which means you can't just strtok() because it will include the space. Therefore, you need to remove the spaces before printing. I attempted to achieve this by strtok() the CSV, then running said token through a

Reading from CSV with fgets() and strtok()

邮差的信 提交于 2020-01-24 13:35:10
问题 FULL DISCLOSURE: SCHOOL ASSIGNMENT I've been working on some code to pull data from a CSV and display it, however I keep encountering this error I can't seem to overcome. The code would be pretty simple, but the problem is that our CSV can have spaces before or after the comma, which means you can't just strtok() because it will include the space. Therefore, you need to remove the spaces before printing. I attempted to achieve this by strtok() the CSV, then running said token through a

Implementation of strtok() function

懵懂的女人 提交于 2020-01-24 00:40:13
问题 I need to write my function strtok. Below is my code. The problem is - I can't display result string. In code I use strcpy() and then display new array. Is it possible to display string using just pointer *str ? #include <stdio.h> #include <string.h> char* my_strtok(char* s, char* delm){ char W[100]; int i = 0, k = 0, j = 0; char *ptr; static char *Iterator; ptr = s; if (s == NULL){ s = Iterator; } while (s[i] != '\0'){ j = 0; while (delm[j] != '\0'){ if (s[i] != delm[j]) W[k] = s[i]; else