Passing two-dimensional array via pointer

Question

How do I pass the m matrix to foo()? if I am not allowed to change the code or the prototype of foo()?

void foo(float **pm)
{
    int i,j;
    for (i = 0; i < 4; i++)
        for (j = 0; j < 4; j++)
            printf("%f\n", pm[i][j]);

}

int main ()
{
    float m[4][4];

    int i,j;
    for (i = 0; i < 4; i++)
        for (j = 0; j < 4; j++)
            m[i][j] = i+j;

    foo(???m???);
}

Answer 1

If you insist on the above declaration of foo, i.e.

void foo(float **pm)

and on using a built-in 2D array, i.e.

float m[4][4];

then the only way to make your foo work with m is to create an extra "row index" array and pass it instead of m

...
float *m_rows[4] = { m[0], m[1], m[2], m[3] };
foo(m_rows);

There no way to pass m to foo directly. It is impossible. The parameter type float ** is hopelessly incompatible with the argument type float [4][4].

Also, since C99 the above can be expressed in a more compact fashion as

foo((float *[]) { m[0], m[1], m[2], m[3] });

P.S. If you look carefully, you'll that this is basically the same thing as what Carl Norum suggested in his answer. Except that Carl is malloc-ing the array memory, which is not absolutely necessary.

Answer 2

If you can't change foo(), you will need to change m. Declare it as float **m, and allocate the memory appropriately. Then call foo(). Something like:

float **m = malloc(4 * sizeof(float *));
int i, j;
for (i = 0; i < 4; i++)
{
    m[i] = malloc(4 * sizeof(float));
    for (j = 0; j < 4; j++)
    {
        m[i][j] = i + j;
    }
}

Don't forget to free() afterwards!