Sorting an array of strings in C using pointers without qsort

Question

So I am a complete beginner to C, and I was trying to sort an array of strings without using qsort. Here is the code:

#include <stdio.h>
#include <string.h>
void sort(char *ar[],int n)
{
    int i,j;
    char temp[10]; 
    for (i=0;i<n-1;i++)
    {
        for (j=0;j<n-i-1;j++)
        {
            if ((strcmp (*(ar+j),*(ar+j+1)))>0)
            {
                strcpy (temp, *(ar+j));
                strcpy (*(ar+j), *(ar+j+1));
                strcpy (*(ar+j+1), temp);
                printf ("%s\n", temp);
            }
        }
    }
    /*printf ("After sorting: \n");
    for (i=0;i<n;i++)
    printf ("%s\n", *(ar));*/
}
int main()
{
    int i,n;
    char* ar[]={"ghi","def","abc"};
    n = sizeof(ar)/sizeof(*ar);
    printf ("Before sorting: \n");
    for (i=0;i<n;i++)
    printf ("%s\n", *(ar+i));
    sort (ar,n);
    printf ("After sorting: \n");
    for (i=0;i<n;i++)
    printf ("%s\n", *(ar+i));
    return 0;
}

However, it only prints the strings before sorting. What am I doing wrong here?

Some programmer dude · Accepted Answer · 2020-05-07 05:39:20Z

3

You have undefined behavior in your sorting function.

There are two reasons:

You use an array of pointers, and each pointer is pointing to a literal string;
You use strcpy to copy contents between the strings.

In C attempting to modify a literal string is undefined behavior. Literal strings are in essence read-only. Note that they are not constant, even through it's always recommended to use const pointers.

You have two possible way of solving this:

Use an array of arrays instead:
```
char ar[][10] = { ... };
```
Then the contents of the strings are modifiable, and you can use strcpy.

Or swap the pointers instead:

char *temp = ar[j];
ar[j] = ar[j + 1];
ar[j + 1] = temp;

answered May 7, 2020 at 5:39

Some programmer dude

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Comments

Daniel Giger · Accepted Answer · 2020-05-07 05:52:23Z

0

Your main problem is that you're using strcpy() with string literals, which isn't allowed. If you swap the pointers instead, your algorithm works perfectly. Here's an updated version of sort():

void sort(char *ar[],int n)
{
    int i,j;
    char* temp; 
    for (i=0;i<n-1;i++)
    {
        for (j=0;j<n-i-1;j++)
        {
            if (strcmp (ar[j], ar[j + 1] ) > 0)  // more readable with indexing syntax
            {
                temp = ar[j];
                ar[j] = ar[j+1];
                ar[j+1] = temp;
                printf ("%s\n", temp);
            }
        }
    }
}

answered May 7, 2020 at 5:52

Daniel Giger

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Comments

tangyiheng · Accepted Answer · 2020-05-07 06:08:52Z

I think you'd better use strdup() to copy a string instead of using strcpy, because the former string will store in heap while the latter string will store in stack which is easily lost after your calling sort().

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

void sort(char *array[], int start, int end) {
    for (int i = start; i <= end - 1; i++) {
        for (int j = i + 1; j <= end; j++) {
            // swap if array[i] > array[j]
            if (strcmp(array[i], array[j]) > 0) {
                char *tmp = strdup(array[j]);
                array[j] = strdup(array[i]);
                array[i] = strdup(tmp);
            }
        } 
    }
}

int main(int argc, char *argv[]) {
    if (argc < 2) {
        fprintf(stderr, "Usage: %s <string> ...\n", argv[0]);
        exit(EXIT_FAILURE);
    }

    sort(argv, 1, argc - 1);            // sort

    for (int i = 1; i < argc; i++) {
        printf("%s ", argv[i]);
    }
    printf("\n");

    exit(EXIT_SUCCESS);
}

$ ./sort_string_array ghi def abc

abc def ghi

or you could just swap between pointers instead of copying one string to another string.

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