Swapping strings in an array of strings

Code:

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

void swap(char* s1, char* s2)
{
    char* temp = s1;
    s1 = s2;
    s2 = temp;
}

int main()
{
    char *str_array[4];

    char buffer[128];

    int i;
    for (i = 0; i < 4; i++)
    {
        printf("Enter string %d: ", i + 1);
        scanf("%s", buffer);

        int length = strlen(buffer); //get the length of the inputed string
        ++length; //add a space for the NULL charactor at the end

        //allocate space at i in the str_array for the word
        str_array[i] = malloc(length * sizeof(char));

        //copy the word into the array
        strcpy(str_array[i], buffer);
    }

    //swap first word and second word
    swap(&str_array[0], &str_array[1]);

    //display words
    for (i = 0; i < 4; i++)
        printf("%s\n", str_array[i]);

    //free allocated memory
    for (i = 0; i < 4; i++)
        free(str_array[i]);

    return 0;
}

If I enter: car boat plane train I want to to swap car and boat. So it would then display boat car plane train.

On the line where I call the swap function i get:

Code:

warning: passing arg 1 of `swap' from incompatible pointer type

What is the correct way to go about swapping strings in an array? Thank you for any help.

You want to swap strings. &str_array[0] is not a string. str_array[0] is.

Actually, I think that the problem is that this function has no net effect:

Code:

void swap(char* s1, char* s2)
{
    char* temp = s1;
    s1 = s2;
    s2 = temp;
}

What you are doing here is swapping the values of the local pointers s1 and s2, but the variables in the caller remain unchanged. What you should do is to have char** parameters instead:

Code:

void swap(char** s1, char** s2)
{
    char* temp = *s1;
    *s1 = *s2;
    *s2 = temp;
}

Once you make this change, your original code would be correct.

Thank you laser light. That worked great.