How do I change this to pointer notation?
Code:#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; }
int addarray(int arr[],int size) changed to int addarray(int *arr, int size)
sum += arr[n]; changed to sum+=*(arr + n)
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.Originally Posted by Dominiko
Last edited by Sir Galahad; 01-26-2018 at 07:18 PM.
Main returns int, not void.
And you could try
sum += *arr++;
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
Thank you everybody for the help
Code:#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; }
Last edited by christophergray; 01-28-2018 at 03:37 PM.
