$a = 5; var_dump($a++); After running the console is 5?

From what I understand, When you run ($a++) it returns $a first and if you were to then var_dump($a) afterwards, it would now be 6.

So: <?php $a = 5; var_dump($a++); var_dump($a); ?>

will return: int(5) int(6)

Let's look at sequentially (the way the computer sees it) :

• first : $a = 5;
  You assign the value 5 to the variable $a.
• then : var_dump($a++); the var_dump() function displays the values of the arguments you pass it to. In our case, the argument is $a++. The expression $a++ does two things, in this order (the order is important) :
• it returns the variable $a
• then it increments the value of $a by one. So, since $a = 5, $a++ returns 5 and then var_dump($a++) returns 5. After that, the value of $a is incremented. You could easily check that if you wrote : var_dump($a++); var_dump($a); You'd see it doesn't show the same value the second time, because $a has been incremented after the first var_dump function was called but before the second one was called.

Does it help you to understand what's happening ?

cheers,

Raphaël

Hi, a++ returns the value of a, and then adds one to the value of a. ++a adds one to the value of a, and then returns the value of a.

cheers,

raphaël

The answer is 5, not 6! And I don't understand why?

with var_dump($a++) you first dump the $a variable and after showing that variable it adds 1 to $a. with var_dump(++$a) it will first add 1 to $a and show it.