Printing an array. (Using a for-loop)

Code:

#include <stdio.h>

void display_array(int array[], int c);


int main(void)
{
	int c, k, array[20];
	
	for(c = 0, k = 0; c < 20; c++)
	{
		printf("Enter a whole number, to end enter: -1. ");
		scanf("%d", &k);
		
		if(k == -1)
		    break;
		    
		array[c] = k;
	}
	
	display_array(array, c);
	
	getch();
}


void display_array(int array[], int c)
{
	for(c = 0; c < 20; c++)
	{
		if(array[c] == )
		    break;
		
		printf("%d\n", array[c]);
	}
}

Now, the problem I get is when a user exits, leaving me with a partially-filled array. Let's say they enter values for 0-8, so 9-19 are left blank, when I run the loop it prints random values for 9-19.

Is there a way to write an if statement to check for whether or not a value has been entered?

Thanks in advance,
~Alan

Why not simply create a counter variable, increment it with each entered value and use it as a loop counter in your function? May not be the best solution, but the first coming to my mind.

Originally Posted by dennis.cpp

Why not simply create a counter variable, increment it with each entered value and use it as a loop counter in your function? May not be the best solution, but the first coming to my mind.

This what you had in mind?

Code:

#include <stdio.h>

void display_array(int array[], int c);


int main(void)
{
	int c, k, array[20];
	
	for(c = 0, k = 0; c < 20; c++)
	{
		printf("Enter a whole number, to end enter: -1. ");
		scanf("%d", &k);
		
		if(k == -1)
		    break;
		    
		array[c] = k;
	}
	
	display_array(array, c);
	
	getch();
}


void display_array(int array[], int c)
{
	int k;
	for(k = 0; k < c; k++)
	{
		printf("%d\n", array[k]);
	}
}

Yea, kind of. I would have done it a bit differently, but a) seems to hit the idea there and b) I'm a beginner too, so both of us could be way off the track here.

Originally Posted by dennis.cpp

Yea, kind of. I would have done it a bit differently, but a) seems to hit the idea there and b) I'm a beginner too, so both of us could be way off the track here.

Well it worked, so w/e.

Learning something is so not about w/e.