Some error in PHP function output I cannot understand

Question

Have a 'function use' lesson problem. The code inside the function (when it's put outside the function; returns an array) seems correct, but the function outputs NULL. Here's what it looks. Point where I'm doing wrong, please.

function getDivisors($num)
       {
         for ($i=1; $i<=$num; $i++)
           {
             if ($num % $i == 0)
               {
                 $divisors[] = $i;
               }
           }
        }

I suppose the output to the array is an incorrect, but... that's still unsure.

Looks like you forgot to return the array from the function — ADyson
– ADyson, Commented Sep 30, 2021 at 13:55
Thank you very much! The function is a part of an exercise, which demands the use of it in function getCommonDivisors($num1,$num2) like this: $result[] = array_intersect(getDivisors($num1), getDivisors($num2)); So the whole thing breaks out with NULL as the variables. — Max Shkel
– Max Shkel, Commented Oct 1, 2021 at 5:41

Atika · Accepted Answer · 2021-09-30 14:13:16Z

1

Your function returns null because you didn't return your array.

Modify your function like this :

function getDivisors($num) {
  for ($i=1; $i<=$num; $i++) {
     if ($num % $i == 0) {
        $divisors[] = $i;
     }
  }
  return $divisors;
}

answered Sep 30, 2021 at 14:13

Atika

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1 Comment

Max Shkel Over a year ago

My great thanks. The topic appeared to have lots of pitfalls for me.

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