Quick question about passing 2d arrays

Suppose I have this code:

Code:

#include <iostream>

void foo(int a[][3]);
int main()
{
    int num[3][3] = {{1,2,3},{4,5,6},{7,8,9}};
   //size of array calculated in here
    foo(num);
    
    
    std::cin.get();
    
    
}
void foo(int a[][3])
{
    std::cout<<a[2][1];
    return;
    
}

What if I don't know that the number of columns is going to be 3 for the two dimensional array? I just know in the function prototype that I need to receive a two dimensional array and I have another variable that will be used for the size of the array? So it would look more like this:

Code:

#include <iostream>

void foo(int a[][], int size);
int main()
{
    int num[3][3] = {{1,2,3},{4,5,6},{7,8,9}};
    //size calculated in here
    foo(num);
    
    
    std::cin.get();
    
    
}
void foo(int a[][], int size)
{
    std::cout<<a[2][1];
    return;
    
}

I have a program where the size of the puzzle is determined by another file (crossword puzzle) so the program isn't very generic when everytime I want to do another puzzle I have to hard code what the size of the puzzle will be in the function prototype/function definition when it can be easily calculated from the files. Thanks.

That is exactly one of the main advantages of STL containers over arrays.

http://www.cprogramming.com/tutorial/stl/vector.html

Good link: http://c-faq.com/aryptr/ary2dfunc3.html

It would work better like this

void foo(int **a, size_t cols, size_t rows);

that way you don't have any guesswork to do about the size of the array and you're doing things right.

[edit] double pwnt. Read up!

Thanks guys.

> void foo(int **a, size_t cols, size_t rows);
NO NO NO!!!
You can't pass a 2D array to a function expecting a pointer to a pointer. Your compiler should warn you, and if it doesn't you need a better one.