c - C 中 printf 的不逻辑总线错误

Question

当我编译并运行代码时，在打印“starting”后立即出现总线错误。发生的情况如下:

bash-3.2$ ./remDup
开始
总线错误

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

void removeDups(char* str) 
{
    int len = strlen(str);
    int i = 0;

    for (i = 0; i < len; i++) {
        char a = str[i];
        int k = i + 1;
        int c = 0;
        int j = 0;

        for (j = k; j < len; j++) {
            if (a != str[j]) {
                str[k] = str[j];
                k++;
            } else c++;
        }

        len -= c;
    }

    str[len] = '\0';
}

int main(int argc, const char* argv[] )
{
    char *str1 = "apple";

    printf("%s -> ", str1);
    removeDups(str1);
    printf("%s\n ", str1);

    return 1;
}

Answer 1

如果将字符串定义为:

char *str1 = "apple";

您不得修改内容 - 标准非常清楚这是未定义的行为^(a)。使用:

char str1[] = "apple";

相反，它会给你一个可修改的副本。它在功能上相当于:

char str1[6]; strcpy (str1, "apple");

<小时/>

^(a) C99 6.4.5“字符串文字”，第 6 段指出:

It is unspecified whether these arrays are distinct provided their elements have the appropriate values. If the program attempts to modify such an array, the behavior is undefined.