Thread: returning a 2D array from a function

Hi all. I am very rusty in C so please bear with me. I want to send a function an array and then return a different array. I understand how to send the array to a function, but not to get it back. Is this the best way of passing arrays from functions by initializing the arrays in the main function?

My error that I ended up is this:

error: expression must be a pointer to a
complete object type
another_array[i][j]=cnt++;

What is wrong with this????

Code:

#include <stdio.h>
#include <stdlib.h>

void function(int array[], int another_array[][]);

int main(void)
{
	int i;
	int array[10];
	int another_array[5][5];
	
	//Create test array
	for(i=0;i<10;i++)
	{
		array[i]=i;
	}
	function(array,another_array);
	
	for(i=0;i<5;i++)
	{
		printf("array[%d]=%d\n",i,array[i]);
		printf("another_array[%d]=%d\n",i,another_array[i][1]);
	}
	
	return 0;
}


void function(int array[], int another_array[][])
{
	int i,j,cnt=0;
	
	for(i=0;i<5;i++)
	{
		for(j=0;j<5;j++)
		{
			another_array[i][j]=cnt++;
		}
	}
	
}

Thank you for your time ^^

Add a 5 to the second dimension of the parameter:

Code:

another_array[][5];

Now re-compile and run the program.

Now put the 5 into the first dimension of the parameter, and re-compile, leaving the second dimension just empty brackets: [].

And re-compile.

What do you surmise from this?

IT WORKS. Reason: I have to define the length of its columns? Why only the columns? And why is this not the same for a 1-D array?

Thanks >_< Trying to understand...

I still don't fully grasp this concept... Please bear with me..

So if you pass an array of 5 ints, it is converted to a pointer to an int, but if you pass an array of 10 arrays of 5 ints, it is converted to a pointer to an array of 5 ints.

Passing an array of 10 arrays with 5 ints? You are referring to a 2D array that is a 10x5, right? So you need to put the 5 in the another_array[][5] so that the pointer is going to an int instead of an array since you cannot send arrays?

So I assume that for a 3D array it is the same thing where as if you initialize another_array[5][5][5], you need to send and return like this:

void function(int array[], int another_array[][5][5]);

Thank you! ^^

The second bracket number is the offset the program uses, to find the next row (which is itself, a pointer to it's row):

So the jump from row to row is just:

Code:

for(i=0;i<MAXROWS; i++) {
  for(column=0;j<MAXCOLS; columns++) {

     array[ i * offset ][columns];

  }
}

0, 5, 10, 15, 20, etc. All right where the program needs to jump to, to find the next ponter of the sub array for that row.

Without that offset (five in this case), how would the program know where to go to, in an array, that was in a called function, other than the one that created it?

IT WORKS -- you need not sound so surprised!

You're welcome.