Need help converting to pointer notation

How do I change this to pointer notation?

Code:

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#include <stdio.h>


int addarray(int arr[],int size);


void main() {
  int array1[10] = {1,2,3,4,5,6,7,8,9,10};
  int array2[4] = {0,0,0,0};


  printf("The sum of array1 = %d\n", addarray(array1,10));
  printf("The sum of array2 = %d\n", addarray(array2,4));
}


int addarray(int arr[],int size)
{
    int sum = 0 , n ;
    for( n = 0 ; n < size ; n++ )
    {
        sum += arr[n];
    }
    return sum;
}

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int addarray(int arr[],int size) changed to int addarray(int *arr, int size)

sum += arr[n]; changed to sum+=*(arr + n)

Quote:

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Originally Posted by Dominiko

How do I change this to pointer notation?

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I would say change the prototype as suggested but leave the array indexing inside of the function. I mean if you're just wondering how its done, there you have it, but otherwise there's really no sense in writing code like that.

Main returns int, not void.

And you could try
sum += *arr++;

Thank you everybody for the help

Code:

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#include <stdio.h>
#include <stdlib.h>
static int addarray(int *, int);
int main(void) {
  { int * array1;
    if ((array1 = (int *)malloc(10 * sizeof(int))) == NULL)
      exit(EXIT_FAILURE);
    for (int i = 0; i < 10; ++i)
      *(array1 + i) = i + 1;
    printf("The sum of array1 = %d\n", addarray(array1, 10));
    free(array1);
  }

  { int * array2;
    if ((array2 = (int *)calloc(4, sizeof(int))) == NULL)
      exit(EXIT_FAILURE);
    printf("The sum of array2 = %d\n", addarray(array2, 4));
    free(array2);
  }
}

int addarray(int * arr, int size) {
  int sum = 0;
  for (int n = 0; n < size; ++n, ++arr)
    sum += *arr;
  return sum;
}

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