How to add adjacent array items in c with pointers?

Question

#include <stdio.h>

void add_adjacents() {
  int num1[5] = {1, 2, 3, 4, 5};
  int num2[5] = {10, 20, 30, 40, 50};
  int final[5];

  for (int i=0; i<sizeof(num1); i++) {
    final[i] = num1[i] + num2[i];
  }

  for (int c=0; c<sizeof(final)/sizeof(final[0]); c++) {
    printf("%d\n", final[c]);
  }
}

void main() {
  add_adjacents();
}

So, I did the above without the pointers. But with pointers, here is my attempt: I'm still new to pointers, and I'm playing with different practice problems.

#include <stdio.h>

void add_adjacents() {
  int num1[5] = {1, 2, 3, 4, 5};
  int num2[5] = {10, 20, 30, 40, 50};
  int final[5];


  for (; *num1 != '\0'; *num1++) {
    *final = *num1 + *num2;
  }

  for (int c=0; c<sizeof(final)/sizeof(final[0]); c++) {
    printf("%d\n", final[c]);
  }
}

void main() {
  add_adjacents();
}

*num1++ : num1 is array, not pointer. *num1 != '\0' : 0 is not included in the array of num1 — BLUEPIXY
– BLUEPIXY, Commented Feb 11, 2017 at 12:10
Oh man! Your code makes so much sense !!! And yeah, I have to CREATE a pointer ... makes sense! Thank you so much. — nastypluto
– nastypluto, Commented Feb 11, 2017 at 12:12

Paul Ogilvie · Accepted Answer · 2017-02-11 12:09:47Z

1

The following does the trick:

void add_adjacents() {
  int num1[5] = {1, 2, 3, 4, 5};
  int num2[5] = {10, 20, 30, 40, 50};
  int final[5], c;

  int *n1= num1, *n2=num2, *f=final;

  for (; n1<&num1[5]; ) {
    *f++ = *n1++ + *n2++;
  }

  for (c=0; c<sizeof(final)/sizeof(final[0]); c++) {
    printf("%d\n", final[c]);
  }
}

answered Feb 11, 2017 at 12:09

Paul Ogilvie

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