php.net |  support |  documentation |  report a bug |  advanced search |  search howto |  statistics |  random bug |  login

go to bug id or search bugs for

Doc Bug #60114 operator increment not working as expected
Submitted: 2011-10-22 16:46 UTC Modified: 2011-10-28 12:33 UTC
Votes:1
Avg. Score:3.0 ± 0.0
Reproduced:0 of 1 (0.0%)
From: luuk34 at gmail dot com Assigned:
Status: Not a bug Package: *General Issues
PHP Version: 5.3.8 OS:
Private report: No CVE-ID: None
View Developer Edit
Welcome back! If you're the original bug submitter, here's where you can edit the bug or add additional notes.
If you forgot your password, you can retrieve your password here.
Password:
Status:
Package:
Bug Type:
Summary:
From: luuk34 at gmail dot com
New email:
PHP Version: OS:

 

 [2011-10-22 16:46 UTC] luuk34 at gmail dot com 
Description:
------------
---
From manual page: http://www.php.net/language.operators.increment
---

I think there's an error in the second example:
$foo=($foo++); 

it should assign 0 to $foo,
and after that increment $foo with 1.

But the value of $foo after this line of code is still 0.



Test script:
---------------
<?php
	$foo=0;
	$foo=$foo++;
	print $foo;

	$foo=0;
	$foo=($foo++);
	print $foo;

	$foo=0;
	$foo++;
	print $foo;
?>	

Expected result:
----------------
011

Actual result:
--------------
001

Patches

Pull Requests

History

AllCommentsChangesGit/SVN commitsRelated reports
 [2011-10-22 17:14 UTC] luuk34 at gmail dot com 
The actual result should even be:
111

and not:
011
 [2011-10-25 07:14 UTC] contact at jubianchi dot fr 
For me, it seems like this is the expected behavior : post-incrementation 
operator returns the value and the do the incrementation.

So the first part of the snippet should print 0 as the second exemple.
Only the third par of the snipper should print 1 as the $foo isn't being assigned 
the return value of the incrementation (we only increment the value).

In the manual : $a++	Post-increment	Returns $a, then increments $a by one.
 [2011-10-25 18:53 UTC] luuk34 at gmail dot com 
You mean to say the incremantation is done BEFORE the assignment? (in the case $foo=$foo++ )
 [2011-10-28 12:33 UTC] rquadling@php.net
-Status: Open +Status: Bogus
 [2011-10-28 12:33 UTC] rquadling@php.net 
$foo = $foo++;
and
$foo = ($foo++);

say, take the current value of $foo and treat that value as the result of the expression (in this instance the right hand side of the assignment). Then increment $foo. Then assign 
the value previously determined as the result to $foo, overriding the post-incremented value of $foo.

The parenthesis do nothing here.

As the documentation says in http://docs.php.net/manual/en/language.operators.increment.php, post-increment is "Returns $a, then increments $a by one."

If you were to use ...

<?php
       $foo=0;
       $newfoo=$foo++;
       echo $foo, $newfoo;

       $foo=0;
       $newfoo=($foo++);
       echo $foo, $newfoo;

       $foo=0;
       $foo++;
       echo $foo;
?>

you would see the difference ...

10101
 
PHP Copyright © 2001-2025 The PHP Group
All rights reserved. 		Last updated: Fri Oct 31 10:00:02 2025 UTC