给定一个 ( char *
) 字符串,我想找到所有出现的子字符串并将它们替换为备用字符串。我在 <string.h>
中没有看到任何实现此目的的简单函数.
最佳答案
优化器应该消除大部分局部变量。 tmp 指针用于确保 strcpy 不必遍历字符串来查找 null。 tmp 指向每次调用后结果的结尾。 (请参阅 Shlemiel the painter's algorithm 了解为什么 strcpy 会很烦人。)
// You must free the result if result is non-NULL.
char *str_replace(char *orig, char *rep, char *with) {
char *result; // the return string
char *ins; // the next insert point
char *tmp; // varies
int len_rep; // length of rep (the string to remove)
int len_with; // length of with (the string to replace rep with)
int len_front; // distance between rep and end of last rep
int count; // number of replacements
// sanity checks and initialization
if (!orig || !rep)
return NULL;
len_rep = strlen(rep);
if (len_rep == 0)
return NULL; // empty rep causes infinite loop during count
if (!with)
with = "";
len_with = strlen(with);
// count the number of replacements needed
ins = orig;
for (count = 0; tmp = strstr(ins, rep); ++count) {
ins = tmp + len_rep;
}
tmp = result = malloc(strlen(orig) + (len_with - len_rep) * count + 1);
if (!result)
return NULL;
// first time through the loop, all the variable are set correctly
// from here on,
// tmp points to the end of the result string
// ins points to the next occurrence of rep in orig
// orig points to the remainder of orig after "end of rep"
while (count--) {
ins = strstr(orig, rep);
len_front = ins - orig;
tmp = strncpy(tmp, orig, len_front) + len_front;
tmp = strcpy(tmp, with) + len_with;
orig += len_front + len_rep; // move to next "end of rep"
}
strcpy(tmp, orig);
return result;
}
关于c - 什么功能是用C语言替换字符串中的子字符串?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/779875/