Saving 2D arrays in a pointer array

Question

I have different sized 2D-arrays and want to store all of them into an array of pointer

int test0[][2] = {{0, 2},
                  {1, 3}};
int test1[][3] = {{10, 20, 30},
                  {40, 50, 60},
                  {70, 80, 90}};

int *t = malloc(2*sizeof(int));

t[0] = (int) test0;
t[1] = (int) test1;

free(t);

So now, the pointers adresses of test1 and test2 are stored in t[0] and t[1] respectively

But I can't access them like t[0][0][0]

Is it possible and how would you do it please?

t as declared is a pointer to an integer, not to a 2d array. No wonder the compiler is complaining. Also, maybe you should have listened to it when it lamented that you were trying to assign test0 and test1 to t[0] and t[1] since I'm sure that's the reason you cast them. — Federico klez Culloca
– Federico klez Culloca, Commented Sep 30, 2019 at 12:16
I tried int ***t = malloc(2*sizeof(int**)) but it won't work neither - I had then t[0] = test0 And I'm not sur if int*** is a real type ? — Xiaojiba
– Xiaojiba, Commented Sep 30, 2019 at 12:17
this type of thing only works with jagged arrays. If t[0] is an int**, how would it know the internal array dimensions? — Blaze
– Blaze, Commented Sep 30, 2019 at 12:19

David Ranieri · Accepted Answer · 2019-10-01 06:59:05Z

4

You can not create an array of objects of different type/size, and in your case the arrays differs in the size.

You need something like:

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

int main(void)
{
    int test0[][3] = {{0, 2},
                      {1, 3}};
    int test1[][3] = {{10, 20, 30},
                      {40, 50, 60},
                      {70, 80, 90}};

    int (**t)[3] = malloc(sizeof(*t) * 2);

    t[0] = test0;
    t[1] = test1;

    printf("t[1][1][1] = %d\n", t[1][1][1]);

    free(t);

    return 0;
}

Output:

But notice that you don't need to use dynamic memory (malloc) :

int (*t[2])[3]; // array 2 of pointer to array 3 of int

will also work.

edited Oct 1, 2019 at 6:59

answered Sep 30, 2019 at 12:25

David Ranieri

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3 Comments

Xiaojiba Over a year ago

Thanks, but I don't understand how it works in the malloc, so t is like t[][3][3]

David Ranieri Over a year ago

Are you sure? are you using the exact same code that I'm positng?

Xiaojiba Over a year ago

I must have made a mistake, the warning is no more. Thanks

Vlad from Moscow · Accepted Answer · 2019-09-30 12:29:21Z

1

If you have an array like this

int test0[][2] = { /*...*/ };

then a pointer to elements of the array looks the following way

int ( *p )[2] = test0;

Or

int test1[][3] = { /*...*/ };
int ( *p )[3] = test1;

and you can access elements of the array using the pointer the same way as you do with the array that is for example p[i][j].

Pay attention to that this code snippet

t[0] = (int) test0;
t[1] = (int) test1;

is unsafe and does not make sense. The size of a pointer can be greater than the size of the type int. And moreover the objects t[0] and t[1] are not pointers. They are integers.

You could define an array of pointers of the type void * and then explicitly cast each element of the array to the required type.

For example

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

int main(void) 
{
    enum { N0 = 2, N1 = 3 };
    int test0[N0][N0] = 
    { 
        { 0, 2 }, 
        { 1, 3 }
    };

    int test1[N1][N1] = 
    {
        { 10, 20, 30 },
        { 40, 50, 60 },
        { 70, 80, 90 }
    };

    void **t = malloc( 2 * sizeof( void * ) );
    t[0] = test0;
    t[1] = test1;

    for ( size_t i = 0; i < N0; i++ )
    {
        for ( size_t j = 0; j < N0; j++ )
        {
            printf( "%d ", ( ( int ( * )[N0] )t[0] )[i][j] );
        }
        putchar( '\n' );
    }

    putchar( '\n' );

    for ( size_t i = 0; i < N1; i++ )
    {
        for ( size_t j = 0; j < N1; j++ )
        {
            printf( "%d ", ( ( int ( * )[N1] )t[1] )[i][j] );
        }
        putchar( '\n' );
    }

    putchar( '\n' );

    free( t );

    return 0;
}

answered Sep 30, 2019 at 12:29

Vlad from Moscow

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2 Comments

Xiaojiba Over a year ago

In int ( *p )[3] = test1; are the parenthesis mandatory ? Because I tried something similar to this !

Vlad from Moscow Over a year ago

@Newtonus This a declaration of a pointer to elements of the type int[3]. Without the parentheses int *p[3] is an array of objects of the type int *. So these are two different types.

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