Question

I made a code to swap two strings:

void swap (char *a, char *b)
{
    char *t = a;
    a = b;
    b = t;
}

int main()
{
    char * strings[2];
    strings [0] = "luck!";
    strings [1] = "good ";
    swap (strings[0], strings[1]);
    printf( "%s %s\n",strings[0], strings[1]);
    return 0;
}

And it fails. What i have trouble understanding is when i call swap() i pass two pointers. Both pointers point to the first character of their assigned arrays. I then created a temporary pointer inside the function and perform basic switch. What is flawed here? I really want to understand why this approach is wrong?

Answer 1

You are switching the parameters of the function, which are local to the function scope. When your function is executed, the parameters (a, of type char*, and b, of type char*) are passed by value, put on the stack, and the function is executed. The parameters are modified and then popped of the stack without effect.

To make a difference, you need to pass references to the parameters:

void swap (char **a, char **b)
{
    char *t = *a;
    *a = *b;
    *b = t;
}

and then call with:

swap (&strings[0], &strings[1]);

You now pass pointers to individual array elements in strings, which is in main’s stack segment and thereby persists past the context of swap.