C function to swap two chars in a char[] segfaults, despite not working with string literals

Question

I've verified in GDB that the program crashes on the *(a) = *(b) line. This does not make sense to me.

In my main function I allocated a 5 bytes for the char* string. I pass two pointers to swap, one is string offset by sizeof(char) and the other is the pointer to string. These pointers are copied to swap()'s call stack. The 5 bytes I allocated earlier should still be on the stack so swap() should have no problem with dereferencing and writing to those locations on stack, right?

  int main(int argc, char* argv[])
  {
      char *string = "abcd";
      swap((string+1), string);
      printf("%s\n",string);
      return 0;
  }

  void swap(char *a, char *b)                                                                                                                                                                       
  {
       if(!a || !b)
           return;

       char temp = *(a);
       *(a) = *(b);
       *(b) = temp;
   }

Solution 1

char *string = "abcd";

is a pointer to a string literal and string literals are immutable in C. Modifying a string literal invokes undefined behavior.

Change the declaration of string to:

char string[] = "abcd";

to fix you program. Here string is an array, initialized with the content of a string literal and is modifiable.

Solution 2

String literals like "abcd" are read only, you must not change them. Use

char string[] = "abcd";

instead.

Comments

[zwol]

You can initialize a char[] with a regular string literal and it'll still be writable. You don't have to write out the { 'a', 'b', 'c', 'd' }. (Unless you really don't want the terminating NUL, which does occasionally come up.)

[Rafael]

crazy enough, but I had no idea this was the case in C, thanks.