Question

Please provide me the proper solution of this script with explanation:

 $a = 5;
 $c = $a-- + $a-- + --$a - --$a;
 echo $c;

What will be the value of $c = 10; Why?

Answer 1

++ and -- produce the same end result – incrementing or decrementing the variable – whether applied before of after the variable name, the difference comes when it is used as part of a larger statement.

Consider this:

$a = 5;
$a--;
echo $a; // 4

$a = 5;
--$a;
echo $a; // 4

So you see, they produce the same end result – $a get decremented by one. I’m sure this is what you were expecting.

However:

$a = 5;
echo $a--; // 5
echo $a; // 4

$a = 5;
echo --$a; // 4
echo $a; // 4

In this example, $a is still decremented after the operation, but the order in which the decrement happens and the value is used is different. For $a-- the value is used before the decrement, and for --$a the value is used after.

So for your example code:

$a = 5;    //  Operations in order of occurence:
$c = $a--  //  $c = 5;          $a = 5 - 1  == 4;
   + $a--  //  $c = 5 + 4 == 9; $a = 4 - 1  == 3;
   + --$a  //  $a = 3 - 1 == 2; $c = 9 + 2  == 11;
   - --$a; //  $a = 2 - 1 == 1; $c = 11 - 1 == 10; 
echo $c;   //  10