c - 为什么这个 strncpy() 实现会在第二次运行时崩溃？

Question

为什么这个 strncpy() 实现在第二次运行时崩溃，而第一次运行正常？

strncpy

Copy characters from string Copies the first n characters of source to destination. If the end of the source C string (which is signaled by a null-character) is found before n characters have been copied, destination is padded with zeros until a total of n characters have been written to it.

No null-character is implicitly appended at the end of destination if source is longer than n (thus, in this case, destination may not be a null terminated C string).

char *strncpy(char *src, char *destStr, int n)
{
    char *save = destStr; //backing up the pointer to the first destStr char
    char *strToCopy = src; //keeps [src] unmodified

    while (n > 0)
    {
        //if [n] > [strToCopy] length (reaches [strToCopy] end),
        //adds n null-teminations to [destStr]
        if (strToCopy = '\0') 
            for (; n > 0 ; ++destStr)
                *destStr = '\0';

        *destStr = *strToCopy;
        strToCopy++;
        destStr++;
        n--;

        //stops copying when reaches [dest] end (overflow protection)
        if (*destStr == '\0')
            n = 0; //exits loop
    }

    return save;
}

/////////////////////////////////////////////

int main()
{
    char st1[] = "ABC";
    char *st2;
    char *st3 = "ZZZZZ";
    st2 = (char *)malloc(5 * sizeof(char));


    printf("Should be: ZZZZZ\n");
    st3 = strncpy(st1, st3, 0);
    printf("%s\n", st3);

    printf("Should be: ABZZZZZ\n");
    st3 = strncpy(st1, st3, 2);
    printf("%s\n", st3);

    printf("Should be: ABCZZZZZ\n");
    st3 = strncpy(st1, st3, 3);
    printf("%s\n", st3);

    printf("Should be: ABC\n");
    st3 = strncpy(st1, st3, 4);
    printf("%s\n", st3);

    printf("Should be: AB\n");
    st2 = strncpy(st1, st2, 2);
    printf("%s\n", st2);

    printf("Should be: AB\n");
    st2 = strncpy(st1, st2, 4);
    printf("%s\n", st2);
}

Answer 1

你得到一个段错误是因为

char *st3 = "ZZZZZ";

目标是一个字符串文字。字符串字面值不能被修改，它们通常存储在写保护的内存中。所以当你打电话时

strncpy(st1, st3, n);

对于 n > 0，您试图修改字符串文字并导致崩溃(不一定，但通常)。

在复制循环中，您忘记了取消引用 strToCopy

if (strToCopy = '\0')

并写了 = 而不是 ==，所以 strToCopy 被设置为 NULL，导致进一步取消引用strToCopy 调用未定义的行为。