Question

I found a piece of code (from one of our developer) and I was wondering why the output of this is 2?

<?php
  $a = 1;
  $a = $a-- +1;
  echo $a;

thanks

Answer 1

I’ll give my explanation a whirl. We’re talking about a variable referencing some value off in the system.

So when you define $a = 1, you are pointing the variable $a to a value 1 that’s off in memory somewhere.

With the second line, you are doing $a = $a-- + 1 so you are creating a new value and setting that to $a. The $a-- retrieves the value of the original $a, which is 1 and adds 1 to make 2 and creates that value somewhere else in memory. So now you have a variable $a which points to 2 and some other value 1 off in memory which along the way decremented to 0, but nothing is pointing at it anymore, so who cares.

Then you echo $a which points to your value of 2.

Edit: Testing Page