Pass By Reference Multidimensional Array With Unknown Size

Question

How to pass by reference multidimensional array with unknown size in C or C++?

EDIT:

For example, in main function I have:

int main(){
    int          


        

Answer 1

H2CO3's solution will work for C99 or a C2011 compiler that supports VLAs. For C89 or a C2011 compiler that doesn't support VLAs, or (God forbid) a K&R C compiler, you'd have to do something else.

Assuming you're passing a contiguously allocated array, you can pass a pointer to the first element (&a[0][0]) along with the dimension sizes, and then treat it as a 1-D array, mapping indices like so:

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

  for (i = 0; i < rows; i++)
  {
    for (j = 0; j < cols; j++)
    {
      a[i * rows + j] = some_value();
    }
  }
}

int main( void )
{
  int arr[10][20];

  foo( &arr[0][0], 10, 20 );
  ...
  return 0;
}

This will work for arrays allocated on the stack:

T a[M][N];

and for dynamically allocated arrays of the form:

T (*ap)[N] = malloc( M * sizeof *ap );

since both will have contiguously allocated rows. This will not work (or at least, not be guaranteed to work) for dynamically allocated arrays of the form:

T **ap = malloc( M * sizeof *ap );
if (ap)
{
  size_t i;
  for (i = 0; i < M; i++)
  {
    ap[i] = malloc( N * sizeof *ap[i] );
  }
}

since it's not guaranteed that all the rows will be allocated contiguously to each other.

Answer 2

This is a sort of comment to the good answer of @John Bode

This will not work (or at least, not be guaranteed to work) for dynamically allocated arrays of the form:

But this variant will:

T **ap = malloc( M * sizeof *ap );
if (ap) return NULL;     ---> some error atention
if (ap)
{
  ap[0] = malloc( M * N * sizeof *ap[i] );
  if (ap[0]) { free(ap); return NULL;}     ---> some error atention
  size_t i;
  for (i = 1; i < M; i++)
  {
    ap[i] = ap[0] + i * N;
  }
}

After use :

free(ap[0]);
free(ap);

for T being int you call foo exactly als for the array int ap[M][N];

  foo( &ap[0][0], M, N);

since you guaranteed that all the rows are allocated contiguously to each other. This allocation is a litter more efficient.

Answer 3

Simply put, you can't. In C, you can't pass by reference, since C has no references. In C++, you can't pass arrays with unknown size, since C++ doesn't support variable-lenght arrays.

Alternative solutions: in C99, pass a pointer to the variable-length array; in C++, pass a reference to std::vector<std::vector<T>>.

Demonstration for C99:

#include <stdio.h>

void foo(int n, int k, int (*arr)[n][k])
{
    int i, j;
    for (i = 0; i < n; i++) {
        for (j = 0; j < k; j++) {
            printf("%3d ", (*arr)[i][j]);
        }
        printf("\n");
    }
}

int main(int argc, char *argv[])
{
    int a = strtol(argv[1], NULL, 10);
    int b = strtol(argv[2], NULL, 10);

    int arr[a][b];
    int i, j;
    for (i = 0; i < a; i++) {
        for (j = 0; j < b; j++) {
            arr[i][j] = i * j;
        }
    }

    foo(a, b, &arr);

    return 0;
}

Demonstration for C++03:

#include <iostream>
#include <vector>
#include <cstdlib>
#include <ctime>

void foo(std::vector < std::vector < int > > &vec)
{
    for (std::vector < std::vector < int > >::iterator i = vec.begin(); i != vec.end(); i++) {
        for (std::vector<int>::iterator j = i->begin(); j != i->end(); j++) {
            std::cout << *j << " ";
        }
        std::cout << std::endl;
    }
}

int main(int argc, char *argv[])
{
    int i = strtol(argv[1], NULL, 10);
    int j = strtol(argv[2], NULL, 10);

    srand(time(NULL));

    std::vector < std::vector < int > > vec;
    vec.resize(i);
    for (std::vector < std::vector < int > >::iterator it = vec.begin(); it != vec.end(); it++) {
        it->resize(j);
        for (std::vector<int>::iterator jt = it->begin(); jt != it->end(); jt++) {
            *jt = random() % 10;
        }
    }

    foo(vec);

    return 0;
}

Answer 4

John Bode's explanation is very good, but there is a little mistake: it should be

i * cols + j

instead of

i * rows + j

Answer 5

If you really want references, then it's only in C++.

En example of a two-dimensional int array passed by reference

void function_taking_an_array(int**& multi_dim_array);

But the reference doesn't have any advantage, so simply use :

void function_taking_an_array(int** multi_dim_array);

I would advice you to use a container to hold your array.