Question

when declaring the two dimensional array

int random[height][width];

and then using it in a function

void populate(int random[height][width], int x, int y)

gives the error variable-size type declared outside of any function.
I know I’m doing something wrong, and that its something small. I just have a bad memory…

Answer 1

I’m going to step up right now and tell you that multidimensional arrays are not worth the brain effort in C or C++. You’re much better off using single-dimensional arrays (or, better yet, standard containers) and writing an indexing function:

inline int index (int x, int y)
{
  return x + y * width;
}

Now for your problem. C++ does not support C99 variable-length arrays. The compiler must know, at compile time, the size of the array. The following, for example, won’t work.

int dim = 4;
int ar[dim];

If dim were const, it would work because the compiler would be able to tell exactly how wide ar should be (because the value of dim wouldn’t change). This is probably the problem you’re encountering.

If you want to be able to change the size at compile-time, you’ll need to do something more difficult, like write a templated reference. You can’t use a pointer for multidimensional arrays because of the way they are laid out in C/C++. A templated example might look like the following aberration:

template <int Width, int Height>
void populate(int (&(&random)[Width])[Height], int x, int y);

This is ugly.

For run-time, you’ll need to use new to allocate data, or use a container type.