Pointer to a pointer function ---> how to return a pointer

Question

I have this pointer to pointer function that takes two arguments and it is suppose to return pointer to the array. But what is the proper way to return pointer to the array in this case? Here is my code:

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

int **multiTable (unsigned int xs, unsigned int ys)
{
    unsigned int i, j;

    int **table = malloc( ys * sizeof(*table)); 
          // should I put malloc(ys * xs * sizeof(*table))?

    for(i = 0; i < ys; i++)
    {
        for(j = 0; j < xs; j++)
        {
           table[i][j] = (j+1) * (i+1);
        }

    }

    return table;  //what would be the proper way to return table? Is this ok?

    free(**table); //by the way, do I need this? 
}

int main()
{
    int sum = 0;
    sum = multiTable2(3,2);

    return 0;
}

what would be the proper way to return table? Is what I written the return table; ok or should it be like return *table or even return **table? And also, do I need to free the table?

You allocate table, but you don't allocate table[x], leading to *undefined behavior, which you also have in that free call. — Some programmer dude
– Some programmer dude, Commented Dec 13, 2014 at 11:18
Then your assignment in the main function is wrong as well. All in all, your compiler should be screaming at you! — Some programmer dude
– Some programmer dude, Commented Dec 13, 2014 at 11:18
And your free call is really wrong, very much. Does this code compile? — Iharob Al Asimi
– Iharob Al Asimi, Commented Dec 13, 2014 at 11:19
Yes, it does compline. So should I change the malloc to int (*table)[xs] = malloc(ys * sizeof(*table)); — BuffetBiffet
– BuffetBiffet, Commented Dec 13, 2014 at 11:24

Iharob Al Asimi · Accepted Answer · 2014-12-13 11:33:13Z

1

I think this is what you are trying to do.

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

int **multiTable(unsigned int xs, unsigned int ys)
{

    unsigned int i, j;
    int **table;

    table = malloc(ys * sizeof(int *)); /* you need space for ys integer arrays */
    if (table == NULL)
        return NULL;
    for (i = 0 ; i < ys ; i++)
    {
        table[i] = malloc(xs * sizeof(int)); /* you need space for xs integers */
        if (table[i] != NULL) /* did malloc succeed? */
        {
            for (j = 0 ; j < xs ; j++)
                table[i][j] = (j + 1) * (i + 1);
        }
    }
    return table;
}

int main()
{
    int **sum;

    sum = multiTable(3, 2);
    /* do somenthing with sum here */
    if (sum != NULL)
    {
        /* now you must use free, like this */
        unsigned int i;
        for (i = 0 ; i < 2 ; i++)
        {
            if (sum[i] != NULL)
                free(sum[i]);
        }
        free(sum);
    }
    return 0;
}

this could be done, but you need to read more to understand the basics.

Your problem is not

how to return a pointer

but instead how to use malloc and free.

edited Dec 13, 2014 at 11:33

answered Dec 13, 2014 at 11:27

Iharob Al Asimi

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4 Comments

BuffetBiffet Over a year ago

If I would like to use only one malloc, would this work: int (*table)[xs] = malloc(ys * sizeof(*table))

BuffetBiffet Over a year ago

Or int **table = malloc(ys * xs * sizeof(*table))?

Iharob Al Asimi Over a year ago

no, it would be something like int *table = malloc(xs * ys * sizeof(int)) but then you would need to compute the index your self.

Iharob Al Asimi Over a year ago

I mean to get the i,j element you do it this way table[i * columnCount + j].

Armali · Accepted Answer · 2015-07-16 13:24:47Z

0

If I would like to use only one malloc, would this work: int (*table)[xs] = malloc(ys * sizeof(*table))

Yes, it would, if you adjust the function according to Matt McNabb's comment:

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

int (*multiTable(unsigned int xs, unsigned int ys))[]
{
    unsigned int i, j;
    int (*table)[xs] = malloc(ys * sizeof *table);
    for (i = 0; i < ys; i++)
    for (j = 0; j < xs; j++)
       table[i][j] = (j+1) * (i+1);
    return table;
}

int main()
{
    const int xs = 3, ys = 2;
    int (*sum)[xs];
    sum = multiTable(xs, ys);
    int i, j;
    for (i = 0; i < ys; i++, puts(""))
    for (j = 0; j < xs; j++)
       printf("%d", sum[i][j]);
    return 0;
}

answered Jul 16, 2015 at 13:24

Armali

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