How does php evaluate the expression $x+++$x++?

Question

Please, take a look at the following expression:

$x = 20;

echo $x+++$x++; // 41

Why is the answer 41 instead of 43?

Answer 1

Altough I can't find any mention in the PHP docs, I guess PHP evaluates from right, and ++ acts like the same operator in C/C++ (see inc/dec PHP operator docs)

So, the expression:

$x+++$x++

in fact does (evaluated from the right):

1. get the value of x → 20
2. increment the value of x → 21
3. get the value of x → 21
4. increment the value of x → 22
5. sum the values got from 1 and 3 → 41

It is worth mentioning, though, that in this case the same results is obtained even if the expression is evaluated from the left.

Answer 2

$x+++$x++; is

1. Start getting sum of values
2. First value is $x (20)
3. Increment $x (now $x is 21)
4. Second value is $x (21)
5. Get sum of first and second values - 20 + 21 = 41
6. Increment $x (now $x is 22)

Another question to read - What's the difference between ++$i and $i++ in PHP?

Answer 3

The expression

$x+++$x++

is evaluated as

($x++) + ($x++)

The $x++ returns the value of the $x and then increments it by 1. So, this:

$x = 20;
$y = $x++;

You end up with $y == 20 and $x == 21. Now, applying this to your expression:

$x = 20;
($x++) + ($x++);

We get first $x++ returned as 20 while incrementing $x to 21; and the second $x++ returned as 21 while incrementing $x to 22. So:

20 + 21

which evaluates to 41, but note that $x is now set to 22.