Returning dynamic array through void function in C

Question

In my C program, I use a void function with the following arguments: One 2D int array, one int pointer that will be used to create the new dynamic array and a last int pointer which will hold a number of counts that will occur inside the function. So the dynamic array is created in the function using malloc and everything works okay, until I print its elements in main() after calling the function. What I get is rubbish instead of the numbers I should see. Here's the function code:

void availableMoves(int array[][3], int *av, int *counter)
{
    int i, j;
    for (i=0; i<3; i++)
    {
        for (j=0; j<3; j++)
        {
            if (array[i][j] == E)
            {
                printf("%d ", 3*i + j + 1);
                (*counter)++;
            }
        }
    }
    av = (int *) malloc(*counter * sizeof(int));
    if (av == NULL)
    {
        printf("ERROR!");
    }
    else
    {
        for (i=0; i<*counter; i++)
            *(av + i) = 0;
        int pos = 0;
        for (i=0; i<3; i++)
        {
            for (j=0; j<3; j++)
            {
                if (array[i][j] == E)
                {
                    *(av + pos++) = 3*i + j + 1;
                }
            }
        }
    }
}

Just commenting on code: Since you have an error case, you should not modify *counter like that in error case. Use a temp int tmp_count = *counter; variable, and assign it back to *counter only if function succeeded. Alternatively, make it abort(); if malloc fails, or something. Avoid producing "partial" result (av=null but *counter is still modified). — hyde
– hyde, Commented Jan 2, 2013 at 11:24
Try to understand pointer-dereference and indexing. *(av + pos++) = 3*i + j + 1; is the same as av[pos++] = 3*i + j + 1;, but most human readers prefer the second form. Similar for (*counter)++; which could be written as *counter += 1;, avoiding the parentheses. — wildplasser
– wildplasser, Commented Jan 2, 2013 at 11:24
@hyde: You're right, but that was a quick check of malloc, I'm not done with it yet. :) — sotirelisc
– sotirelisc, Commented Jan 2, 2013 at 14:11
@wildplasser: Is there something wrong with the parentheses? — sotirelisc
– sotirelisc, Commented Jan 2, 2013 at 14:11
No, there is nothing wrong with them. But most people tend to reduce the number of parentheses, just because it is easier to read with fewer ((.)(.)). — wildplasser
– wildplasser, Commented Jan 2, 2013 at 14:15

md5 · Accepted Answer · 2013-01-02 11:05:06Z

2

In this function, av is a pointer passed by copy. So when you change the value of your pointer inside the function, the original pointer won't be modified.

There are two possibilities :

use a pointer to pointer (int **av);
return the allocated pointer (return av).

So either:

void availableMoves(int array[][3], int **av, int *counter);

Or:

int *availableMoves(int array[][3], int *av, int *counter)

And the call:

availableMoves(array, &av, &counter);
av = availableMoves(array, av, &counter);

answered Jan 2, 2013 at 11:05

md5

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3 Comments

hyde Over a year ago

Third (IMO most consistent) approach would be to return boolean value: return true when both av and counter have been successfully modified, otherwise return false and leave av and counter unmodified.

md5 Over a year ago

@hyde: Could be indeed a good idea to use the return value otherwise.

sotirelisc Over a year ago

Decided to use the first approach and it seems to be working! Thank you!

BenMorel · Accepted Answer · 2013-12-06 14:58:43Z

2

use double pointer for your dynamic array int **av instead of int *av

void availableMoves(int array[][3], int **av, int *counter)

and into the function change av by *av

edited Dec 6, 2013 at 14:58

BenMorel

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answered Jan 2, 2013 at 11:02

MOHAMED

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