Thread: qsort with array initialized using malloc?

Code:

#include <stdio.h> /* printf */
#include <stdlib.h> /* for malloc */
#include <string.h> /* strcpy */
  
#define WORDSIZE 20
  
int compare2strs( const void *a, const void *b );
void n( char array[][ WORDSIZE ], int N );
void p( char array[][ WORDSIZE ], int N);
  
int main()
{
  
    int N = 10;
    char (*ptr)[WORDSIZE];
  
    ptr = malloc( N * sizeof(*ptr) );
  
    n( ptr, N );
    p( ptr, N);
    return( 0 );
}
int compare2strs( const void *a, const void *b )
{
    char *left = (char *) a;
    char *right = (char *) b;
    return( strcmp( left, right ) );
}
void n( char array[][ WORDSIZE ], int N )
{
    int i;
    for( i = 0; i < N; i++ )
    {
        strcpy( array[ i ], "Cluster!" );
    }
    strcpy( array[3], "try");
    strcpy( array[6], "to");
    strcpy( array[9], "sort");
    strcpy( array[1], "me");
  
    qsort( array, N,
            sizeof( array[ 0 ] ), compare2strs );
  
}
void p( char array[][ WORDSIZE ], int N){
    int i;
  
    for( i = 0; i < N; i++ ){
        printf("%s\n",array[i]);
    }
}

$ gcc foo.c
$ ./a.out 
Cluster!
Cluster!
Cluster!
Cluster!
Cluster!
Cluster!
me
sort
to
try
$

> Our teacher gave us this small piece of example code and I am trying to expand on it to serve my purpose.
The brutal way of assigning the initial pointer, then casting it to pass it to your functions is concerning.

The main issue for it not working was this
> sizeof( array ) / sizeof( array[ 0 ] )
This only works on true arrays which are in scope at the time you use it.
Function parameters (no matter how you declare them) are just pointers.

Your problem has nothing to do with the comparison function.

The problem is that, when arrays are passed to functions, they are passed as pointers. The information about the actual size of the array (as seen by the caller) is lost.

So the trick sizeof(array)/sizeof(array[0]) does not give the number of elements in the array within the function. sizeof(array) will always be the size of a pointer (which is 4 on a typical 32 bit system). You need to pass the array dimensions to your function as separate arguments.

You'll see what I mean with this code.

Code:

#include <stdio.h>

void func(int array[])
{
     int n = (int)(sizeof(array)/sizeof(array[0]));
     printf("Number of elements computed in called function is %d\n", n);
}

int main()
{
     int array[27];
     int n = (int)(sizeof(array)/sizeof(array[0]));
     printf("Number of elements computed in main() is %d\n", n);
     func(array);
}

Despite the calculation of size looking identical, it gives different results depending on where it occurs.