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>* - Kissed strtok() goodbye*/strtok()这个函数大家都应该碰到过,但好像总有些问题, 这里着重讲下它
下面我们来看一个例子：
int main() {
char test1[] = "feng,ke,wei";
char *test2 = "feng,ke,wei";
char *p; p = strtok(test1, ",");
while(p)
{
printf("%s\n", p);
p = strtok(NULL, ",");
}
return 0;
}
运行结果:
feng
ke
wei
但如果用p = strtok(test2, ",")则会出现内存错误,这是为什么呢?是不是跟它里面那个静态变量有关呢? 我们来看看它的原码:
/*** *strtok.c - tokenize a string with given delimiters * * Copyright (c) Microsoft Corporation. All rights reserved. * *Purpose: * defines strtok() - breaks string into series of token * via repeated calls. * *******************************************************************************/
#include#include #ifdef _MT #include #endif /* _MT */
/*** *char *strtok(string, control) - tokenize string with delimiter in control * *Purpose: * strtok considers the string to consist of a sequence of zero or more * text tokens separated by spans of one or more control chars. the first * call, with string specified, returns a pointer to the first char of the * first token, and will write a null char into string immediately * following the returned token. subsequent calls with zero for the first * argument (string) will work thru the string until no tokens remain. the * control string may be different from call to call. when no tokens remain * in string a NULL pointer is returned. remember the control chars with a * bit map, one bit per ascii char. the null char is always a control char. * //这里已经说得很详细了!!比MSDN都好! *Entry: * char *string - string to tokenize, or NULL to get next token * char *control - string of characters to use as delimiters * *Exit: * returns pointer to first token in string, or if string * was NULL, to next token * returns NULL when no more tokens remain. * *Uses: * *Exceptions: * *******************************************************************************/
char * __cdecl strtok ( char * string, const char * control ) { unsigned char *str; const unsigned char *ctrl = control; unsigned char map[32]; int count;
#ifdef _MT _ptiddata ptd = _getptd(); #else /* _MT */ static char *nextoken; //保存剩余子串的静态变量 #endif /* _MT */
/* Clear control map */ for (count = 0; count < 32; count++) map[count] = 0;
/* Set bits in delimiter table */ do { map[*ctrl >> 3] |= (1 << (*ctrl & 7)); } while (*ctrl++); /* Initialize str. If string is NULL, set str to the saved * pointer (i.e., continue breaking tokens out of the string * from the last strtok call) */ if (string) str = string; //第一次调用函数所用到的原串
else #ifdef _MT str = ptd->_token; #else /* _MT */ str = nextoken; //将函数第一参数设置为NULL时调用的余串
#endif /* _MT */
/* Find beginning of token (skip over leading delimiters). Note that * there is no token iff this loop sets str to point to the terminal * null (*str == '\0') */ while ( (map[*str >> 3] & (1 << (*str & 7))) && *str ) str++;
string = str; //此时的string返回余串的执行结果
/* Find the end of the token. If it is not the end of the string, * put a null there. */
//这里就是处理的核心了, 找到分隔符,并将其设置为'\0',当然'\0'也将保存在返回的串中 for ( ; *str ; str++ ) if ( map[*str >> 3] & (1 << (*str & 7)) ) { *str++ = '\0'; //这里就相当于修改了串的内容 ① break; } /* Update nextoken (or the corresponding field in the per-thread data * structure */ #ifdef _MT ptd->_token = str; #else /* _MT */ nextoken = str; //将余串保存在静态变量中,以便下次调用 #endif /* _MT */
/* Determine if a token has been found. */ if ( string == str ) return NULL; else return string; 1. strtok介绍 众所周知，strtok可以根据用户所提供的分割符（同时分隔符也可以为复数比如“，。”） 将一段字符串分割直到遇到"\0". 比如，分隔符=“，” 字符串=“Fred，John，Ann” 通过strtok 就可以把3个字符串 “Fred” “John” “Ann”提取出来。 上面的C代码为
QUOTE:
int in=0; char buffer[]="Fred,John,Ann" char *p[3]; char *buff = buffer; while((p[in]=strtok(buf,","))!=NULL) { i++; buf=NULL; }
如上代码，第一次执行strtok需要以目标字符串的地址为第一参数（buf=buffer），之后strtok需要以NULL为第一参数 (buf=NULL)。指针列p[],则储存了分割后的结果，p[0]="John",p[1]="John",p[2]="Ann"，而buf就变 成 Fred\0John\0Ann\0。
2. strtok的弱点让我们更改一下我们的计划：我们有一段字符串 "Fred male 25,John male 62,Anna female 16" 我们希望把这个字符串整理输入到一个struct， QUOTE:
struct person { char [25] name ; char [6] sex; char [4] age; }
要做到这个，其中一个方法就是先提取一段被“，”分割的字符串，然后再将其以“ ”（空格）分割。
比如： 截取 "Fred male 25" 然后分割成 "Fred" "male" "25"以下我写了个小程序去表现这个过程： QUOTE:
#include<stdio.h> #include<string.h> #define INFO_MAX_SZ 255 int main() { int in=0; char buffer[INFO_MAX_SZ]="Fred male 25,John male 62,Anna female 16"; char *p[20]; char *buf=buffer; while((p[in]=strtok(buf,","))!=NULL) { buf=p[in]; while((p[in]=strtok(buf," "))!=NULL) { in++; buf=NULL; } p[in++]="***"; //表现分割 buf=NULL; } printf("Here we have %d strings\n",i); for (int j=0; j<in; j++) printf(">%s<\n",p[j]); return 0; }
这个程序输出为：
Here we have 4 strings>Fred<>male<>25<>***<这只是一小段的数据，并不是我们需要的。但这是为什么呢？ 这是因为strtok使用一个static（静态）指针来操作数据，让我来分析一下以上代码的运行过程：红色为strtok的内置指针指向的位置，蓝色为strtok对字符串的修改1. "Fred male 25,John male 62,Anna female 16" //外循环2. "Fred male 25\0John male 62,Anna female 16" //进入内循环3. "Fred\0male 25\0John male 62,Anna female 16"4. "Fred\0male\025\0John male 62,Anna female 16"5 "Fred\0male\025\0John male 62,Anna female 16" //内循环遇到"\0"回到外循环6 "Fred\0male\025\0John male 62,Anna female 16" //外循环遇到"\0"运行结束。3. 使用strtok_r在这种情况我们应该使用strtok_r, strtok reentrant. char *strtok_r(char *s, const char *delim, char **ptrptr);相对strtok我们需要为strtok提供一个指针来操作，而不是像strtok使用配套的指针。代码： QUOTE:
#include<stdio.h> #include<string.h> #define INFO_MAX_SZ 255 int main() { int in=0; char buffer[INFO_MAX_SZ]="Fred male 25,John male 62,Anna female 16"; char *p[20]; char *buf=buffer; char *outer_ptr=NULL; char *inner_ptr=NULL; while((p[in]=strtok_r(buf,",",&outer_ptr))!=NULL) { buf=p[in]; while((p[in]=strtok_r(buf," ",&inner_ptr))!=NULL) { in++; buf=NULL; } p[in++]="***"; buf=NULL; } printf("Here we have %d strings\n",i); for (int j=0; jn<i; j++) printf(">%s<\n",p[j]); return 0; }
这一次的输出为：
Here we have 12 strings>Fred<>male<>25<>***<>John<>male<>62<>***<>Anna<>female<>16<>***<让我来分析一下以上代码的运行过程：红色为strtok_r的outer_ptr指向的位置，紫色为strtok_r的inner_ptr指向的位置，蓝色为strtok对字符串的修改1. "Fred male 25,John male 62,Anna female 16" //外循环2. "Fred male 25\0John male 62,Anna female 16"//进入内循环3. "Fred\0male 25\0John male 62,Anna female 16"4 "Fred\0male\025\0John male 62,Anna female 16"5 "Fred\0male\025\0John male 62,Anna female 16" //内循环遇到"\0"回到外循环6 "Fred\0male\025\0John male 62\0Anna female 16"//进入内循环}
原来, 该函数修改了原串.
所以,当使用char *test2 = "feng,ke,wei"作为第一个参数传入时,在位置①处, 由于test2指向的内容保存在文字常量区,该区的内容是不能修改的,所以会出现内存错误. 而char test1[] = "feng,ke,wei" 中的test1指向的内容是保存在栈区的,所以可以修改.
看到这里 大家应该会对文字常量区有个更加理性的认识吧.....