Question

I wanted to implement a reverse function for a null-terminated char* string (null-terminated char string, is that redundant?) and came up with the following solution.

The final swap swap(&tmp, &str) in the reverse function does not have any effect. What’s the reason, how is it done probably?

Remark: I have already reimplemented the reverse function by using only the string itself, a temporary char and indices only, but I am so interested in why this does not work. What haven’t I considered in this pointer matter?

#include <stdlib.h>
#include <stdio.h>

int strlen(char* str)
{
        int len = 0;
        char* ptr = str;
        while (ptr[0] != '\0')
        {
                len++;
                ptr++;
        }

        return len;
}

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


void reverse(char* str)
{
        int len = strlen(str);
        char* tmp = malloc(sizeof(char) * (len + 1));
        int i;
        for (i = 0; i < len; i++)
        {
                tmp[i] = str[len - 1 - i];
        }

        tmp[len] = '\0';
        // printf(tmp); => "wolfrevOkcatS olleH"
        swap(&tmp, &str);
}

int main(int argc, char** argv)
{
        char* str = "Hello Stackoverflow\0";
        reverse(str);
        printf("%s", str);
        return 0;
}

Answer 1

You can just do an in place reverse, and you won’t need that pointer swap you’re doing:

void reverse(char* str, int len)
{ 
        int i;
        for (i = 0; i < len; i++)
        {
                char tmp = str[i];
                str[i] = str[len - 1 - i];
                str[len - 1 - i] = tmp;
        }
}

To answer your question, you need to pass a char** to your reverse function, because you’re changing what it will point to, and pointers are copied by value, so if you want to reflect the change of pointee you need a double pointer.