trouble with qsort

I have written this to test qsort. The array takes the values, but qsort isn't working in my code. I think that there may also be a problem with calloc, becasue I am able to enter more entries than supp;ied by calloc, what do you think?

Code:

/*A program that reads into an array created to hold x amount of ints, then order the array using qsort and finally print it out*/
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>

int compare_doubles(double *v1, double *v2);
int read_into_array(double[]);
void print_array(double[], int);


int
main(int argc, char **argv) {
	double *array;
	int n, i;
	
	printf("Enter how many values to be stored in the array: ");
	scanf("%d", &n);
	
	array = calloc(n, sizeof(double));
	
	for (i = 0; i < n; i++)
	{
		array[i]=0;
	}
	printf("Now put some values into the array to sort: ");
	
	n = read_into_array(array);
	
	qsort(array, n, sizeof(*array), compare_doubles);
	
	print_array(array, n);

	return 0;
}

int
compare_doubles (double *ptr1, double *ptr2) {
	if (ptr1<ptr2)
	{
		return -1;
	}
	if (ptr1>ptr2)
	{
		return 1;
	}
	return 0;
}

int
read_into_array(double A[]) {
	int cntr=0;
	double number;
	
	while (scanf("%lf", &number)==1)
	{
		A[cntr]=number;
		cntr++;	
	}
	return cntr;
}

void print_array(double A[], int n) {
	int i;
	
	for (i=0; i<n; i++) {
		printf("%.3f   ", A[i]);
	}
	printf("\n");
	return;
}

Take a good hard look at your comparison function.
You're not comparing double's, you are in fact comparing pointers!
Surprise surprise, the pointers are already in order, so no swaps take place.
You need to compare *ptr1 < *ptr2 etc.

yes, I changes the function to :

Code:

int
compare_doubles (const void *a, const void *b) {
	double *arg1=(double*)a;
	double *arg2=(double*)b;
	if (*arg1<*arg2)
	{
		return -1;
	}
	if (*arg1>*arg2)
	{
		return 1;
	}
	return 0;
}

now it works

[QUOTE=qubit67;639319]

Code:

double *arg1=(double*)a;
double *arg2=(double*)b;

You don't need the (double *) on the right hand side. A void pointer can always be cast silently. But you should be declaring these as "const double *" not just "double *". Surely you get a warning?

Originally Posted by brewbuck

Originally Posted by qubit67

Code:

double *arg1=(double*)a;
double *arg2=(double*)b;

You don't need the (double *) on the right hand side. A void pointer can always be cast silently. But you should be declaring these as "const double *" not just "double *". Surely you get a warning?

There should't be any warning due to implicit cast - it removes const modifier by force

Originally Posted by vart

There should't be any warning due to implicit cast - it removes const modifier by force

You mean explicit I assume? You're right, I'm so used to NOT explicitly casting for a conversion like this that I forgot it wouldn't warn.