Returning array (pointer) of 2D array in C

Question

A function dynamically creates an int array whose elements are predetermined to be int[2]. Is there any way to have a function assign values to that array and then return it to the caller.

The below code accomplishes this task but throws warning: initialization from incompatible pointer type [enabled by default]

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

int *get_values()
{
    int (*x)[2] = malloc(sizeof(int[2])*3);

    x[0][0] = 1;
    x[0][1] = 2;

    x[1][0] = 11;
    x[1][1] = 12;

    x[2][0] = 21;
    x[2][1] = 22;

    return x;   
}

int main()
{
    int (*x)[2] = get_values();

    int i;
    for (i=0; i!=3; i++)
        printf("x[%d] = { %d, %d }\n", i, x[i][0], x[i][1]);
}

I'm aware of the alternative where you dynamically allocate both dimensions, but this is something that I am curious about.

@PeterA.Schneider Maybe because he doesn't know how to put it in the prototype ;) The correct prototype for your function should be int (*get_values())[2] { ... }. — Holt
– Holt, Commented Apr 6, 2016 at 11:46
I wasn't quite sure either and wanted to check first. But "how do I declare..." was not what the OP asked ;-). — Peter - Reinstate Monica
– Peter - Reinstate Monica, Commented Apr 6, 2016 at 11:52
Btw, I think it's fine to simply cast the pointers around as needed. — Peter - Reinstate Monica
– Peter - Reinstate Monica, Commented Apr 6, 2016 at 11:56

Spikatrix · Accepted Answer · 2016-04-06 11:55:31Z

7

Rather than keep repeating the same clunky syntax it can be helpful to define a typedef in cases like this. This makes it easier to declare the correct return type for get_values:

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

typedef int I2[2];

I2 * get_values(void)
{
    I2 * x = malloc(sizeof(I2) * 3);

    x[0][0] = 1;
    x[0][1] = 2;

    x[1][0] = 11;
    x[1][1] = 12;

    x[2][0] = 21;
    x[2][1] = 22;

    return x;   
}

int main()
{
    I2 * x = get_values();

    int i;
    for (i=0; i!=3; i++)
        printf("x[%d] = { %d, %d }\n", i, x[i][0], x[i][1]);

    free(x);
}

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Recommended reading: Don't repeat yourself (DRY).

edited Apr 6, 2016 at 11:55

Spikatrix

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answered Apr 6, 2016 at 11:42

Paul R

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Comments

2501 · Accepted Answer · 2016-04-06 11:47:50Z

4

And this is how it looks without a typedef:

int (*get_values(void))[2]
{  
    return NULL;   
}

Pretty unreadable.

Notice in that function definition, if you replace get_values(void) with x you get: int (*x)[2], which is exactly what the pointer definition looks like.

answered Apr 6, 2016 at 11:47

2501

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13 Comments

Spikatrix Over a year ago

@2501 But it compiles for me without any warnings when omitting the 2 (GCC).

2501 Over a year ago

@CoolGuy Apparently in this case you can omit the 2, however if the returned type doesn't match with the assigned type, the behavior is undefined. Omitting the size also removes warnings that will warn you about the type discrepancy. In any case, there is no good reason to omit it, even you if are allowed to.

Lundin Over a year ago

@2501 Yeah int(*get_values((int*)[2]))[2] is totally trivial reading :)

Peter - Reinstate Monica Over a year ago

@CoolGuy I'm wondering what meaning gcc bestows on int (*get_values(void))[]. In particular, what value it thinks get_values()+1 should have then.

2501 Over a year ago

@PeterA.Schneider It is a pointer to an incomplete type. It can't be dereferenced. I'm not sure if int(*)[] is compatible with int(*)[2], but gcc thinks it is.

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