Question

Though we declare a function with an integer array, we pass address of the array to the function. In the case of simple integers it gives error if we pass address we get pointer conversion error. But how its possible in case of an array

#include<stdio.h>
void print_array(int array[][100],int x, int y);
main()
{
    int i,j,arr[100][100];
    printf("Enter the array");
    for(i=0;i<2;i++)
    {
        for(j=0;j<2;j++)
        {
            scanf("%d",&arr[i][j]);
        }
    }
    print_array(arr,i,j);

}

void print_array(int array[][100],int x,int y)
{
    int i,j;
    printf("\nThe values are\n");
    for(i=0;i<x;i++)
    {
        for(j=0;j<y;j++)
        {
            printf("%d",array[i][j]);
        }
    }
}

My question is even though our function is declared as one with integer array as first parameter (here) we are passing array address when we call the function. How does it function?

Answer 1

Your are passing the array, not its address.
arr is an int[][] array
(in fact it is pretty the same as &(arr[0]), which is a pointer to (the address of) the first line of your array. In C, there is no practical difference between an array and the corresponding pointer, except you take it’s address with the & operator.)

Edit: Ok, just to make me clear:

#include <stdio.h>

int fn(char p1 [][100], char (*p2)[100])
{
  if (sizeof(p1)!=sizeof(p2))
    printf("I'm failed. %i <> %i\n",sizeof(p1),sizeof(p2));
  else
    printf("Feeling lucky. %i == %i\n",sizeof(p1),sizeof(p2));
}

int main()
{
  char arr[5][100];
  char (*p)[100]=&(arr[0]);
  fn(arr, arr);
  fn(p, p);
  return 0;
}