How to set the size of an array through function arguments?

Question

C++ / C#

Let's say there is a one-dimensional array arr (like m*n elements). It is passed to the function, as well as the length and width of the two-dimensional array created from it. In the end something like this:

int * foo(int * arr, int m, int n) {
    int tempArray[m][n];
    ...
    ...
}

But it does not allow you to set the dimensions in this way ("m and n must have a constant value). How then can this be done?

Answer 1

Allocate memory dynamically: Only then do not forget to free it.

Answer 2

But it does not allow you to set the dimensions in this way ("m and n must have a constant value). How then can this be done?

C99 allows. Use C99 already, 18 years have passed since its adoption.
A related question/answer about passing a multidimensional array directly through function arguments: How do I put an n - dimensional array into a function argument?

Answer 3

In memory, the array will still be one-dimensional, even crack;) N-dimensionality is needed for ease of access.

Answer 4

template<size_t N, size_t M>
int* foo () {
    int tempArray[ N ][ M ];
    //...
}

// use case
foo<2, 3>();

Of course, the calculation must be in compile-time.