反向字符串是seg faulting

Question

I've written a "reverse string" exercise as a preparation for my upcoming interviews, however, when I try to run the test on several items of my array, I am getting a segfault.

Basically if I compile the code below with -DWORKS, cstr with the string "zip it shrimp" is reversed as expected. However, if I leave it will compile the code that uses the array of pointers but fails on the first iteration (ie cstr[i] where i = 0):

*s = *e;

I seem to recall that we can't change a pointer directly, but I can't seem to think of a way to work around this. Any help understanding what exactly is leading to the fault and how I can fix it and understand this better so I don't repeat it again, would be greatly appreciated. Whole code below.

#include <stdio.h>

char *str[] = { "zip it shrimp", "", "a", "ab", "abc" };
char cstr[] = "zip it shrimp";

void reverse(char *str)
{       
        char *s = str;
        char *e = s + strlen(s) - 1;

        while (s < e) {
                char c = *s;
                *s = *e;
                *e = c;
                s++;
                e--;
        }
}

int main(int argc, char **argv)
{
        int i;
#ifdef WORKS
        reverse(&cstr);
        printf("%s\n", cstr);
#else
        for (i = 0; i < 4; i++) {
                reverse(str[i]);
                printf("%s\n", str[i]);
        }
#endif
        return 0; 
}       

Answer 1

When reverse the array of pointers, you are modifying the string literals, which is undefined behaviour . You can't modify any of strings in:

char *str[] = { "zip it shrimp", "", "a", "ab", "abc" };

Copy the strings into a modifiable memory locations and then reverse them. One way is to use strdup() .

  for (i = 0; i < 4; i++) { 
            char *p=strdup(str[i]);
            if (!p) { /* handle error */ }
            reverse(p);
            printf("%s\n", p);
            free(p);
    }

If you simply want to print them in reverse, you can print them without doing any copy at all. But that depends on how you want implement it.

Answer 2

If you don't explicitly reserve mutable space for string literals (section 6.4.5 in the language spec) as you do for cstr , they are not guaranteed to be assigned mutable storage.

See also Is modification of string literals undefined behaviour according to the C89 standard? .

C11, draft spec N1516, section 6.4.5.7 states:

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.

C99, draft spec WG14/N1256, section 6.4.5.6 states:

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.

The compiler is free to take such literals and merge them with literals of other modules. If you were to allow modifications of such strings, you might affect string literals from other moduls. In Linux, such literals are stored in .rodata sections (read-only data) and the loader will ensure that they are not writeable-- which leads to the segmentation fault that you are observing.