How does the compare function in qsort work?
I found this sample code online, which explains how the qsort
function works. I could not understand what the compare function returns.
#include "stdlib.h"
int values[] = { 88, 56, 100, 2, 25 };
int cmpfunc (const void * a, const void * b) //what is it returning?
{
return ( *(int*)a - *(int*)b ); //What is a and b?
}
int main(int argc, _TCHAR* argv[])
{
int n;
printf("Before sorting the list is: \n");
for( n = 0 ; n < 5; n++ ) {
printf("%d ", values[n]);
}
qsort(values, 5, sizeof(int), cmpfunc);
printf("\nAfter sorting the list is: \n");
for( n = 0 ; n < 5; n++ ) {
printf("%d ", values[n]);
}
return 0;
}
int cmpfunc (const void * a, const void * b) //what is it returning?
{
return ( *(int*)a - *(int*)b ); //What is a and b?
}
Is equivalent to:
int cmpfunc (const void * a, const void * b) //what is it returning?
{
// qsort() passes in `void*` types because it can't know the actual types being sorted
// convert those pointers to pointers to int and deref them to get the actual int values
int val1 = *(int*)a;
int val2 = *(int*)b;
// qsort() expects the comparison function to return:
//
// a negative result if val1 < val2
// 0 if val1 == val2
// a positive result if val1 > val2
return ( val1 - val2 );
}
What a
and b
are is clearly stated in the documentation for qsort
: these are pointers that point to array elements that have to be compared.
Comparison function in this case knows that the array elements are of type int
. So it casts the void *
pointers to int *
type and performs a tri-state comparison of the pointed int
values by subtracting the second from the first.
It will work properly for your set of values, but in general case using subtraction for tri-state comparison is a bad programming practice, since it is prone to overflow. Also, the function in your sample code unnecessarily casts away the constness of the pointed values.
A better alternative would be
int cmpfunc(const void *a, const void *b)
{
const int *A = a, *B = b;
return (*A > *B) - (*A < *B);
}