Swapping strings WITHOUT using double pointers

Question

This is actually just a small part of the original assignment I have in algorithms and data structure. At some point I'm supposed to write a function that swaps two elements in an array of strings.

The prototype we have to use is:

void swap(char *a, char *b);

To me it makes much more sense passing double pointers to this function (if swapping strings). In fact, I don't know how to make their prototype work without using double pointers. Am I missing something or is this wrong?

I've done my research on this a while ago and there are many answered questions here, I'm just wondering whether the person who was designing this assignment made a mistake.

You can only use that swap function if a and b point to the same amount of memory (or an excessive amount of memory). E.g. char s1[100] = "hello", s2[100] = "world"; swap(s1, s2) — user3386109
– user3386109, Commented May 6, 2018 at 21:38
@user3386109 ah, you're right. Good point. Seems we have to make the assumption that enough memory is allocated at each pointer to contain either C-string — Patrick Roberts
– Patrick Roberts, Commented May 6, 2018 at 21:41
The accepted answer conflicts with the problem statement "write a function that swaps two elements in an array of strings" ... swapping elements is not the same as swapping strings. — RamblinRose
– RamblinRose, Commented May 6, 2018 at 22:47

machine_1 · Accepted Answer · 2018-05-06 22:17:48Z

3

You can swap two strings by creating a temporary char array of length equal to either one of the two strings, and by using strcpy().

Of course, you need to make sure that swapping does not overflow one of the two strings/arrays. Here is a sample example:

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


void swap(char *a, char *b)
{
    char *tmp = malloc(strlen(a) + 1);

    strcpy(tmp, a);
    strcpy(a, b);
    strcpy(b, tmp);

    free(tmp);
}

int main(void)
{
    char str_a[50] = "Hello world";
    char str_b[50] = "What's up";

    swap(str_a, str_b);

    puts(str_a);
    puts(str_b);

    return 0;
}

Another approach is to use a loop with a single char variable, thus avoiding dynamic memory allocation:

void swap(char *a, char *b)
{
    char tmp;

    while (*a && *b)
    {
        tmp = *a;
        *a++ = *b;
        *b++ = tmp;
    }

    if (!*a)
    {
        size_t n = 0;
        while (b[n])
        {
            *a++ = b[n++];
        }
    }
    else if (!*b)
    {
        size_t n = 0;
        while (a[n])
        {
            *b++ = a[n++];
        }
    }
    *a = '\0';
    *b = '\0';
}

edited May 6, 2018 at 22:17

answered May 6, 2018 at 21:53

machine_1

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6 Comments

machine_1 Over a year ago

@melpomene Because I use visual studio :)

melpomene Over a year ago

That makes no sense. Do you mean you have it configured to use C++?

melpomene Over a year ago

You only need one char of scratch space. You don't really need to malloc a whole string.

machine_1 Over a year ago

@melpomene Yes. The C compiler inside visual studio is very old, and disgusting.

machine_1 Over a year ago

@melpomene Do you mean, I can simply create a char variable, and use it in a loop to swap one character at a time?

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