no matter what I do my code would result the same

Question

I wonder why I am getting the result as 4

<?php
function minusNum($aNum){
    $result=$aNum-1;
    return $result;
}

$theNum=4;
minusNum($theNum);
echo $theNum;
?>

Answer 1

Your minusNum function returns a new value, it doesn't manipulate the inputted $aNum . You should use that returned value. Eg:

$result = minusNum($theNum);
echo $result;

Answer 2

You need to capture the result from your function.

  <?php

  function minusNum($aNum){
      $result = $aNum - 1;
      return $result;
  }

  $theNum = 4;

  $theNum = minusNum($theNum); 

  echo $theNum;
 ?>

Here you assign the result of the function back to your original variable $theNum .

Although you pass the variable into the function, it is only altered inside the function. The value remains the same outside of the function.

Answer 3

<?php
function minusNum($aNum){
    $result=$aNum-1;
    return $result;
}

$theNum=4;
echo minusNum($theNum);
?>

Echo the function because it returns result but when you echo $theNum you will get the $theNum value which is 4. Moreover the function will grape the value of any value variable that passes through not the variable it self. so you contradicting youself. I hope it helps.