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I'd /ike to know, how to pass pointers to dynamically allocated arrays using function arguments. This function is supposed to allocate array 10x10 (checks skipped for simplicity sake). Is this possible? What am i doing wrong? Thanks in advance.

int array_allocate2DArray ( int **array, unsigned int size_x, unsigned int size_y)
{
    array = malloc (size_x * sizeof(int *));

    for (int i = 0; i < size_x; i++)
        array[i] = malloc(size_y * sizeof(int));

    return 0;
}

int main()
{
    int **array;
    array_allocate2DArray (*&array, 10, 10);
}

2 Answers 2

1

I came across this post when I was facing a similar problem (I was looking for a way to dynamically allocate an array of strings in C). I prefer to return the array pointer from the function. The following worked for me (I adapted it for your array of integers). I arbitrarily set 99 for each value so I could see them printed out in main.

int **array_allocate2DArray(unsigned int size_x, unsigned int size_y)
{
        int i;
        int **arr;

        arr = malloc(size_x*(sizeof(int*)));

        for (i=0 ; i<size_x ; i++){
                arr[i] = malloc(size_y);
                *arr[i] = 99;
        }

        return arr;

}

int main(void)
{
        int i;
        int **arr;

        arr = array_allocate2DArray(10, 10);

        for (i=0 ; i<10 ; i++){
                printf("%d\n", *arr[i]);
                free(arr[i]);
        }
        free(arr);

        return 0;
}
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Comments

0

Try something like this:

int array_allocate2DArray (int ***p, unsigned int size_x, unsigned int size_y)
{
    int **array = malloc (size_x * sizeof (int *));

    for (int i = 0; i < size_x; i++)
        array[i] = malloc(size_y * sizeof(int));

    *p = array;
    return 0;
}


int **array;
array_allocate2DArray (&array, 10, 10);

I used the temporary p to avoid confusion.

1 Comment

why use &array here and not simply pass array in int array_allocate2DArray(int **p,...)

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