Thread: incompatible pointer type error?

Howdy!

I'm learning C and ran into the error above with the note:
"expected 'int *' but argument is of type 'int **' "
What exactly did I mess up and how do I fix it? Here is my full code:

#include <stdio.h>


void sum(int a[], int n, int *even_elt_sum_ptr, int *odd_elt_sum_ptr);

Code:

int main (void)
{
	int n=10, *even_elt_sum_ptr , *odd_elt_sum_ptr , odd, even;
	int a[10] = {1,2,3,4,5,6,7,8,9,10};
	
	even_elt_sum_ptr = 0;
	odd_elt_sum_ptr = 0;
	
	sum(a, n, &even_elt_sum_ptr, &odd_elt_sum_ptr);
	
	even = *even_elt_sum_ptr;
	odd = *odd_elt_sum_ptr;
	
	printf("the even sum is %d\n", even);
	printf("the odd sum is %d\n", odd);
	
	return 0;
}


void sum(int a[], int n, int *even_elt_sum_ptr, int *odd_elt_sum_ptr)
{
	int x;
	
	for (x=0;x<n;x++){
		if (a[x] / 2 == 0){
			*even_elt_sum_ptr = *even_elt_sum_ptr + a[x];
		}
		else
			*odd_elt_sum_ptr = *odd_elt_sum_ptr + a[x];
	}
}

The code is intended to take an array of 10 integers and return the sum of all of the odd elements and a separate sum for all of the even elements. As I am still learning I am open to any comments/suggestions on improvements that you see. Thanks in advance!

Originally Posted by GReaper

Inside main, remove even_elt_sum_ptr and odd_elt_sum_ptr and pass the addresses of even and odd directly to the function.

Also, if you want to find out if a number is even or odd, you need modulo, not divide.

Thank you so much that fixed it, I did not realize that I was trying to pass the address of a pointer which made no sense.