Passing a 2 dimension array to a function

I am just learning C++ programming and want to pass a 2
dimension array to a function that will oprint this array. My problem is the function needs to know the element size subscript. How can I get this parameter to the function without hardcoding the size in the function?

Just pass the row/col values to the function:

Code:

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void print_array(int **array, int row, int col);

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Quote:

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Originally posted by Monster

Code:

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void print_array(int **array, int row, int col);

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you don't want to pass a pointer to a pointer to the datatype of the array. You want to pass just a pointer to the datatye of the array. IE

Code:

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void DoSomething( Datatype* Array, unsigned int NumRows, unsigned int NumCols );

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and then manually calculate element offset from row and column values within the function

If you happen to know the number of columns before-hand, you can do something like

Code:

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void DoSomething( Datatype Array[][NumCols], unsigned int NumRows ); // Where NumCols is a constant

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or if you knew both the number of rows and columns before hand you could do

Code:

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void DoSomething( Datatype (&Array)[NumRows][NumCols] ); // Where NumRows and NumCols are constants

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I know OOP's method, but I don't know moster's, so how does it work?

Quote:

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Originally posted by ammar
I know OOP's method, but I don't know moster's, so how does it work?

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It doesn't -- at least without typecasting

Quote:

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Originally posted by Polymorphic OOP
It doesn't -- at least without typecasting

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It does work:

Code:

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#include <iostream>

void print_array(int **array, int row, int col)
{
  for(int r = 0; r < row; r++)
  {
      for(int c = 0; c < col; c++)
        cout << array[r][c] << " ";
      cout << endl;
  }
}

int main(void)
{
  int rows, cols, **array;

  cout << "Enter number of rows:";
  cin >> rows;
  cout << "Enter number of columns:";
  cin >> cols;

  // allocate memory
  array = new int *[rows];

  for(int i = 0; i < rows; i++)
      array[i] = new int[cols];

  // fill array
  for(int r = 0; r < rows; r++)
      for(int c = 0; c < cols; c++)
        array[r][c] = r + c;

  print_array(array, rows, cols);

  // delete memory
  for(int i = 0; i < rows; i++)
      delete [] array[i];

  delete array;

  return 0;
}

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Quote:

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Originally posted by Monster
It does work

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No, it doesn't. The only reason you got that to work was because you tailored the situation to work with an array of pointers rather than a multidimensional array. They are very different things.

Quote:

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Originally posted by Polymorphic OOP
No, it doesn't. The only reason you got that to work was because you tailored the situation to work with an array of pointers rather than a multidimensional array. They are very different things.

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I know they are different things but you can use them the same way (as a multidimensional array).

vVv's code is a good example of the debate. vVv declares a clear multidimensional array, foobar, and passes it to foo using a multilevel pointer, p, and uses that multilevel pointer as an array in the function. So, is p a pointer or an array? I say: Who cares! As long as you use it like an array, it's an array. Bottom line: you can achieve the desired result with any of the above techniques. Take your pick.