Thread: Two dimensional arrays and pointers

I'm not understanding what's happening here.


It seems that I'm feeding the function in the wrong way

Code:

#define LEN 4

int sum_two_dimensional_array(const int a[][LEN], int n)
{   
    const int *p;
    int sum = 0;
    
    for (p = &a[0][0]; p < &a[n-1][LEN-1]; p++)
        sum += *p;
    
    return sum;
}

int main(void)
{
    int arr[3][LEN] = {{1, 2, 3, 4}, {1, 2, 3, 4}, {1, 0, 3, 2}};
    
    printf("The sum is: %d\n", sum_two_dimensional_array(arr, 3));
    
    return 0;
}

It's just an exercises from a book. modify the function int sum_two_dimensional_array(const int a[][LEN], int n);
Using pointers to eliminate the use of two for loops, variables i & j and the [ ] (which I dont know how can I write the for loop for twodim array without using the [ ] )
Any explanation it's appreciate

Maybe this is a better approach:

Code:

#include <stdio.h>

#define LEN 4

// n = number of "lines", m = number of "columns".
// p pointer to the first element.
int sum_two_dimensional_array ( const int *p, 
                                int n, int m )
{
  int sum = 0;

  // Points to next address beyond the array boundary.
  const int *q = p + m*n;

  while ( p < q )
    sum += *p++;

  return sum;
}

int main ( void )
{
  int arr[][LEN] = {
    {1, 2, 3, 4}, 
    {1, 2, 3, 4}, 
    {1, 0, 3, 2}
  };

  printf ( "The sum is: %d\n", 
    sum_two_dimensional_array ( (const int *)arr, 
                                3, 4 ) );

  return 0;
}