本文共 3213 字，大约阅读时间需要 10 分钟。
先来看一个程序：
#include#include int main(){ char str[] = "a = 1\nb = 2\nc = 3\nd = 4\ne = 5\nf = 6"; char left[100] = {0}; char right[100] = {0}; char delims[] = "\n"; char *result = strtok(str, delims); while(result != NULL) { printf( "%s\n", result); sscanf(result, "%s = %s", left, right); printf( "%s:%s\n", left, right); result = strtok(NULL, delims); } return 0;}
程序结果为：
a = 1
a:1 b = 2 b:2 c = 3 c:3 d = 4 d:4 e = 5 e:5 f = 6 f:6当然，下面这种情况就没法用sscanf来分割了（其实，需要对sscanf函数的返回值进行判断）：
#include#include int main(){ char str[] = "a=1\nb=2\nc=3\nd=4\ne=5\nf=6"; char left[100] = {0}; char right[100] = {0}; char delims[] = "\n"; char *result = strtok(str, delims); while(result != NULL) { printf( "%s\n", result); sscanf(result, "%s = %s", left, right); printf( "%s:%s\n", left, right); result = strtok(NULL, delims); } return 0;}
所以，要想其他办法来解析，用strtok呗，程序如下：
#include#include void splitIntoTwoParts(char *org, char sep, char *left, char *right, int n){ char delims[2] = {0}; delims[0] = sep; char *result = strtok(org, delims); strncpy(left, result, n - 1); while(result != NULL) { result = strtok(NULL, delims); strncpy(right, result, n - 1); return; }}int main(){ char str[] = "a=1"; char left[100] = {0}; char right[100] = {0}; splitIntoTwoParts(str, '=', left, right, 100); printf("%s\n", left); printf("%s\n", right); return 0;}
结果为：
a
1
可见，splitIntoTwoParts函数没什么问题（当然，为了简便起见，我没有考虑异常情况）
strtok的诡异就在于，下面程序是有问题的：
#include#include void splitIntoTwoParts(char *org, char sep, char *left, char *right, int n){ char delims[2] = {0}; delims[0] = sep; char *result = strtok(org, delims); strncpy(left, result, n - 1); while(result != NULL) { result = strtok(NULL, delims); strncpy(right, result, n - 1); return; }}int main(){ char str[] = "a=1\nb=2\nc=3\nd=4\ne=5\nf=6"; char left[100] = {0}; char right[100] = {0}; char delims[] = "\n"; char *result = strtok(str, delims); while(result != NULL) { printf( "%s\n", result); splitIntoTwoParts(result, '=', left, right, 100); printf( "%s:%s\n", left, right); result = strtok(NULL, delims); } return 0;}
结果为：
a=1 a:1居然没有读到bcdef, 我晕，这个问题花了我1个小时，去百度搜一下就知道，你就明白了，原来，不能嵌套使用strtok函数。strtok好蹩脚啊。上述程序可以改为：
#include#include void splitIntoTwoParts(char *org, char sep, char *left, char *right, int n){ char delims[2] = {0}; delims[0] = sep; char *result = strtok(org, delims); strncpy(left, result, n - 1); while(result != NULL) { result = strtok(NULL, delims); strncpy(right, result, n - 1); return; }}int main(){ char str[] = "a=1\nb=2\nc=3\nd=4\ne=5\nf=6"; char left[100] = {0}; char right[100] = {0}; char delims[] = "\n"; char *result = strtok(str, delims); char line[20][100] = {0}; int times = 0; while(result != NULL) { printf( "%s\n", result); strncpy(line[times++], result, 100 - 1); result = strtok(NULL, delims); } int i = 0; for(i = 0; i < times; i++) { splitIntoTwoParts(line[i], '=', left, right, 100); printf("%s:%s\n", left, right); } return 0;}
结果为：
a=1 b=2 c=3 d=4 e=5 f=6 a:1 b:2 c:3 d:4 e:5 f:6注意，strtok函数还会引起原串的变化，真他妈的蹩脚的设计。
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