CODEWITHSUNDEEP
phparrays
gtilflm
gtilflm

Reputation: 1475

Factors of a number in an array (PHP function)

I'm probably missing something simple here, but I have this function for finding the factors of a number.

 function factor($n){
 $factors_array = array();
 for ($x = 1; $x <= sqrt(abs($n)); $x++)
 {
    if ($n % $x == 0)
    {
        $z = $n/$x; 
        array_push($factors_array, $x, $z);
       }
   }
   return $factors_array;
 }

I then want to do something like this...

factor(120);
print_r($factors_array);

This give me nothing though. Any ideas on where I'm going wrong?

Upvotes: 0

Views: 2566

Answers (2)

maxton
maxton

Reputation: 392

You aren't assigning the variable to the return value of the function. As far as the PHP interpreter is concerned, $factors_array only exists if you're inside the factor() function. Try this:

$factors_array = factor(120);
print_r($factors_array);

Then you can reuse $factors_array in other areas of the code.

Have a look at this page for an explanation of why this happens.

Upvotes: 2

user1646111
user1646111

Reputation:

just try this:

print_r(factor(120));

Because you can't access $factors_array; outside the function, this is called scope of variable, usually, variables that defined inside function are not available outside, also, variables that are defined outside function are not available inside function...

Read more Variable scope ¶

Upvotes: 0

Related Questions