c - 用其他子字符串替换字符串的子字符串时出现段错误

Question

我有字符串“{”1”:“[4,11,14,19,20,18,27]”}“。我想把它改成"{\"1\":\"4,11,14,19,20,18,27\"}"。

下面是我的代码:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char *replace (char *this, char *withthat, char *inthis) {
    char *where = inthis;

    while ((where = strstr(where, this))) {
        memcpy(where, withthat, strlen(withthat));
        memmove(where+strlen(withthat),where+strlen(this), strlen(where+strlen(this))+1);
    }
    return inthis;
}

int main(void) {
    char string[] = "{&quot;1&quot;:&quot;[4,11,14,19,20,18,27]&quot;}";
    printf("%s\n", replace("&quot;", "\\\"", string));
    printf("%s\n", replace("\"[" , "\"", string));
    printf("%s\n", replace("]\\" , "\\", string));
    printf("%s\n", replace("{" , "\"{", string));
    printf("%s\n", replace("}" , "}\"", string));
    return 0;
}

我收到最后两次替换调用的错误。我的运算符(operator)是 {\"1\":\"[4,11,14,19,20,18,27]\"}{\"1\":\"4,11,14,19,20,18,27]\"}{\"1\":\"4,11,14,19,20,18,27\"}段错误

我尝试使用 gdb，但无法找到错误的根本原因。它在某种程度上与 memcopy 有关，但无法理解。如果有人能帮助我，那就太好了。提前致谢。

Answer 1

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char *replace (char *old, char *new, char *buff) {
    char *ptr;
    size_t oldlen = strlen(old);
    size_t newlen = strlen(new);

    for(ptr=buff; ptr = strstr(ptr, old); ptr += newlen) {
        memmove(ptr+newlen, ptr+oldlen, strlen(ptr+oldlen)+1);
        memcpy(ptr, new, newlen);
    }
    return buff;
}

int main(void) {
    char string[1234] = "{&quot;1&quot;:&quot;[4,11,14,19,20,18,27]&quot;}";
    printf("%s\n", replace("&quot;", "\\\"", string));
    printf("%s\n", replace("\"[" , "\"", string));
    printf("%s\n", replace("]\\" , "\\", string));
    printf("%s\n", replace("{" , "\"{", string));
    printf("%s\n", replace("}" , "}\"", string));

    return 0;
}

最后两个替换“{}”包含其自身。这会导致原始字符串在同一位置重新扫描，重新匹配+重新替换。于无穷。 ptr+=newlen 避免了这种情况。