How to return a 2D array?

Question

I was searching for an efficient way to create a 2D array and return it, usually I make an array of pointers, allocate memory for each pointer, and return an **ptr. I was looking for another way to do it because then I have to free every single memory allocated.

I was reading online that you can allocate a 2D array like this: int (*arr)[n] = malloc( sizeof *arr * i ); In which ptr is a pointer, pointing to the adress of arr[n].

When I try to return arr, from the following function: int *array_of_smallest(int count);
I get: warning: assignment to ‘int *’ from incompatible pointer type ‘int (*)[n]’ [-Wincompatible-pointer-types]

If I initialise another pointer and point it to array, then return it, I'm only returning a 1D array.

I think I'm mixing diferent concepts and confusing myself.

With an array of pointers I don't have a problem, but I wanted a simplier way to create a 2D array, also easier to free, but I'm not being able to return that array from my function.

Answer 1

You declared a pointer of the variable modified type type int ( * )[n].

int (*arr)[n] = malloc( sizeof *arr * i );

That is the variable n used in the declarator is not an integer constant expression.

In this case the function should have the return type void *. And it can be declared like

void * array_of_smallest(int count)
{
    int (*arr)[n] = malloc( sizeof *arr * i );
    //...
    return arr;
} 

In the caller you will write

int ( *arr )[n] = array_of_smallest( count );

In this declaration the value of n must be the same as used within the function.

If to use an integer constant expression like

int (*arr)[2] = malloc( sizeof *arr * 2 );

then the function declaration will look like

int ( * array_of_smallest(int count) )[2];

Or you can introduce a typedef name before the function declaration like

typedef int Array[2];

and in this case the function will look like

Array * array_of_smallest(int count);