Making an array of even numbers out of another array

[alun]

What I have to do is make a function that will make an array of even numbers out of another array of ints that is inputted by the user. Here is the function that I have so far... Sorry if it is absolutely horrible. I suck with arrays so badly, and i'm pretty new at it.

void distributeEven(int array[])
{
	int counter = 0,
	    array2[10],
	    arrayCounter = 0;

	for (counter; counter < 10; counter++)
	{
		if (array[counter] % 2 == 0)
		{
			array2[arrayCounter] = array[counter];
		}
		arrayCounter++;
	}
}

I know its horrible and probably has many problems, one of them being that I don't know how to make it return values into the array when i'm calling it.

I thank you for any help that you can afford me

[Peconet009]

This code might help you:
Program to initialize array by even numbers starting from 2 and display sum of all elements - C++ Programming Examples and Tutorials

[Shantanu]

This is the program (in C) for creating arrays of even nos :

#include <stdio.h>

int main()
{
        int i,j,count=0;
        printf("Enter the size of an array: ");
        scanf("%d",&i);
        int array[i],array_even[i];
        printf("Enter the elements of an array :\n ");
        for(j=0;j<i;j++)
        {
                scanf("%d",&array[i]);
                if(array[i]%2==0)
                {
                        array_even[count]=array[i];
                        count++;
                }

        }

        printf("Array of even elements(from given array):\n");
        for(j=0;j<count;j++)
        printf("%d\n",array_even[j]);

        return 0;
}

The output of the program looks like this :

shantanu@{Just Do It !}:~/C_progs$ vi forum_array.c 
shantanu@{Just Do It !}:~/C_progs$ gcc forum_array.c 
shantanu@{Just Do It !}:~/C_progs$ ./a.out
Enter the size of an array: 10
Enter the elements of an array :
 15
69
83
26
48
3
128
9
742
588
Array of even elements(from given array):
26
48
128
742
588
shantanu@{Just Do It !}:~/C_progs$

I hope this is what you need & it address your query !