Passing a 2d array of unkown size

[dannyzimbabwe]

#include <iostream>

using namespace std;

void show(int **array, int size)
{
    for (int i = 0; i < size; i++)
    {
        for (int j = 0; j < size; j++)
        {
            cout <<array[i][j];
        }
        cout << endl;
    }
}

int main()
{
    int array[4][4] =
    {
        { 1, 1, 1, 1 },
        { 1, 0, 0, 1 },
        { 1, 0, 0, 1 },
        { 1, 1, 1, 1 }
    };

    show(array, 4);

    return 0;
}

error: cannot convert `int (*)[4]' to `int**' for argument `1' to `void show(int**, int)'

I need to pass 2d n by n arrays to a function. However, the sizes will be different but the function will do the same thing. How would I go about doing this?

Thanks for any help and suggestions

[anon]

You could definitely pass a one-dimensional "view" of the array.

#include <iostream>

using namespace std;

void show(int *array, int row_count, int col_count)
{
    for (int i = 0; i < row_count; i++)
    {
        for (int j = 0; j < col_count; j++)
        {
            cout <<*array;
            ++array;
        }
        cout << endl;
    }
}

int main()
{
    int array[4][4] =
    {
        { 1, 1, 1, 1 },
        { 1, 0, 0, 1 },
        { 1, 0, 0, 1 },
        { 1, 1, 1, 1 }
    };

    show(array[0], 4, 4);

    return 0;
}

[JVene]

You've encountered one of the reasons it's better to use a vector, where size is part of the vector class - or, better still (more flexible) to pass an iterator to the vector.

[hk_mp5kpdw]

In such cases, you'd need to at least be able to specify the last dimension of the array for what you were trying to actually work.

void show(int *array[4], int size)

[dannyzimbabwe]

Thank you all for the help. I will use anon's solution until I can learn to design it better.