How to return a multidimensional static array from a function in c

Question

I want to return a multidimensional matrix from a function.

So I know how to do that through a struct and dynamic allocating. But I'm not sure how to do that with a static multidimensional array.

#include <stdio.h>
static int **function(){
    static int array[5][10];
    return array;
}
int main(void){
    int** test;
    test = function();
    return 0;
}

The Gcc keep emiting this warning:

Warning: returning 'int (*)[10]' from a function with incompatible return type 'int **' [-Wincompatible-pointer-types] return array;

And I want to understand why?

The error message actually contains the actual type you're returning: "returning 'int (*)[10]' ". That's the return type you need. You might want to create a type-alias (using typedef) for it. — Some programmer dude
– Some programmer dude, Commented Jul 13, 2021 at 19:23
As for understanding what's happening, remember that arrays decays to pointers to their first element. That is, when you do return array; then array decay to &array[0]. And since array[0] is an array of ten int elements, the type of &array[0] is "pointer to array of ten int", or int (*)[10]. — Some programmer dude
– Some programmer dude, Commented Jul 13, 2021 at 19:25

Vlad from Moscow · Accepted Answer · 2021-07-13 19:50:26Z

In this return statement

return array;

the array designator array is converted to pointer to its first element of the type int ( * )[10]. But the function return type is int **. There is no implicit conversion from the type int ( * )[10] to the type int **. So the compiler will issue an error.

Pay attention to that the user of the function needs to know the sizes of the array.

So I advise you to declare the function like

static int ( *function( void ) )[5][10]
{
    static int array[5][10];
    return &array;
}

and then in main you can write

int ( *test )[5][10] = function();

dereferencing the pointer test you will get the array.

Here is a demonstrative program.

#include <stdio.h>

enum { M = 5, N = 10 };

static int ( *function( void ) )[M][N] 
{
    static int array[M][N] =
    {
        { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 },
        { 2, 2, 2, 2, 2, 2, 2, 2, 2, 2 },
        { 3, 3, 3, 3, 3, 3, 3, 3, 3, 3 },
        { 4, 4, 4, 4, 4, 4, 4, 4, 4, 4 },
        { 5, 5, 5, 5, 5, 5, 5, 5, 5, 5 }
    };
    
    return &array;
}

int main(void) 
{
    int ( *test )[M][N] = function();
    
    printf( "The size of the array is %zu\n", sizeof( *test ) );
    
    for ( size_t i = 0; i < M; i++ )
    {
        for ( size_t j = 0; j < N; j++ )
        {
            printf( "%d ", ( *test )[i][j] );
        }
        putchar( '\n' );
    }
    
    return 0;
}

The program output is

The size of the array is 200
1 1 1 1 1 1 1 1 1 1 
2 2 2 2 2 2 2 2 2 2 
3 3 3 3 3 3 3 3 3 3 
4 4 4 4 4 4 4 4 4 4 
5 5 5 5 5 5 5 5 5 5

You can introduce an alias for the array type the following way to make the function declaration simpler

#include <stdio.h>

enum { M = 5, N = 10 };

typedef int Array[M][N];

static Array * function( void ) 
{
    static Array array =
    {
        { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 },
        { 2, 2, 2, 2, 2, 2, 2, 2, 2, 2 },
        { 3, 3, 3, 3, 3, 3, 3, 3, 3, 3 },
        { 4, 4, 4, 4, 4, 4, 4, 4, 4, 4 },
        { 5, 5, 5, 5, 5, 5, 5, 5, 5, 5 }
    };
    
    return &array;
}

int main(void) 
{
    Array *test = function();
    
    printf( "The size of the array is %zu\n", sizeof( *test ) );
    
    for ( size_t i = 0; i < M; i++ )
    {
        for ( size_t j = 0; j < N; j++ )
        {
            printf( "%d ", ( *test )[i][j] );
        }
        putchar( '\n' );
    }
    
    return 0;
}

Tom Karzes · Accepted Answer · 2021-07-13 19:24:54Z

1

There is a big difference between a "pointer to a pointer" and a "pointer to an array". In your case, you need a "pointer to an array", which requires knowing the second dimension of the array. Try the following:

#include <stdio.h>

static int (*function(void))[10] {
    static int array[5][10];
    return array;
}

int main(void) {
    int (*test)[10];
    test = function();
    return 0;
}

Then from main you can access it as test[i][j].

answered Jul 13, 2021 at 19:24

Tom Karzes

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1 Comment

Some programmer dude Over a year ago

You might want to show a version using a type-alias as well, as that will make thins very much simpler.

0___________ · Accepted Answer · 2021-07-13 20:58:28Z

0

You can also use void * function.

static void *function(void) 
{
    static int array[5][10];
    return array;
}

int main(void) 
{
    int (*test)[10] = function();
}

answered Jul 13, 2021 at 20:58

0___________

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