Question

void displayMatrix(int **ptr)
{
  printf("%d %d \n",**ptr,*(*ptr+1));
  *ptr++;
  printf("%d %d \n",**ptr,*(*ptr+1));
}

Is it a correct way to pass 2×2 array to function ?

displayMAtrix(array); 

Answer 1

If you have a 2D array with automatic storage duration (e.g. int matrix[2][2];), then no, this is not the correct way to pass a 2D array.

int f ( int, char ** )
{
  int matrix[2][2];
  displayMatrix(matrix);
}

To be standard compliant (there are other ways to make it work, but this is the standard, recommended way), you need to declare displayMatrix() as:

void displayMatrix ( int matrix[][2] );

You must declare the size of each dimension (possibly excluding the first). The reason behind this lies with the way 2D arrays are stored in memory. Wikipedia has a decent article on row-major order explaining the layout.

Alternate storage type

If you’re allocating large matrices (e.g. for storing images), you’ll usually end up with double pointers because you’ll be allocating the memory differently. In this case, you usually have a 1D array of pointers with each item storing the pointer to a 1D array representing rows (or columns).

In that case, you would get something like:

// this function is over-simplified.  it may leak memory if any
// but the first `new` throws `std::bad_alloc`.
int ** new_matrix ( int m, int n )
{
  int ** matrix = new int*[m];
  for (int i = 1; (i < m); ++i ) {
    matrix[i] = new int[n];
  }
  return (matrix);
}

void displayMatrix ( int ** matrix, int m, int n );

int main ( int, char ** )
{
  int **const matrix = new_matrix(2, 2);
  displayMatrix(matrix, 2, 2);
}