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Bug #66089 The result of post-increment/post-decrement operation is included in result
Submitted: 2013-11-13 14:16 UTC Modified: 2013-11-13 14:46 UTC
From: alexanderhook at gmail dot com Assigned:
Status: Not a bug Package: Variables related
PHP Version: 5.4.21 OS: Windows 7, Ubuntu 12
Private report: No CVE-ID: None
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 [2013-11-13 14:16 UTC] alexanderhook at gmail dot com 
Description:
------------
In some cases, if variable is has participation in expression with post-increment, then the result of post-increment operation is included in the result of the whole expression.

Test script:
---------------
$i = 1;
$sum = $i + ($i++);      //3 , but should be 2
$i = 1;
$sum = 0 + $i + ($i++);  //2 , correct 
$i = 1;
$sum = $i + $i + ($i++); //3 , correct

$i = 1;
$sum = $i + $i--;        // 1
$i = 1;
$sum = 0 + $i + $i--;    // 2 
$i = 1;
$sum = $i + $i + $i--;   // 3



Expected result:
----------------
$i = 1;
$sum = $i + ($i++);      //2
$i = 1;
$sum = 0 + $i + ($i++);  //2
$i = 1;
$sum = $i + $i + ($i++); //3

$i = 1;
$sum = $i + $i--;        // 2
$i = 1;
$sum = 0 + $i + $i--;    // 2 
$i = 1;
$sum = $i + $i + $i--;   // 3

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 [2013-11-13 14:23 UTC] alexanderhook at gmail dot com
-Operating System: Windows, Ubuntu +Operating System: Windows 7, Ubuntu 12
 [2013-11-13 14:23 UTC] alexanderhook at gmail dot com 
Test:

$i = 1;
$sum = $i + ($i++);
assert($sum == 2);

$i = 1;
$sum = $i + $i--;    
assert($sum == 2);
 [2013-11-13 14:29 UTC] derick@php.net
-Status: Open +Status: Not a bug
 [2013-11-13 14:29 UTC] derick@php.net 
Thank you for taking the time to write to us, but this is not
a bug. Please double-check the documentation available at
http://www.php.net/manual/ and the instructions on how to report
a bug at http://bugs.php.net/how-to-report.php

Reading and writing the same variable during an assignment gives undefined behaviour.
 [2013-11-13 14:46 UTC] alexanderhook at gmail dot com 
Excuse me, please check again. The php versions showing different results for these statements:

In php 4.4.9 result is "2":
$i = 1;
echo ($i + ($i++)); 

In php 5.0.5 result is "2":
$i = 1;
echo ($i + ($i++)); 


In php 5.1.5 and higher the result is "3":
$i = 1;
echo ($i + ($i++));
 [2013-11-13 15:35 UTC] nikic@php.net 
@alexanderhook: Yes, this is exactly what Derick told you. For more information see http://us1.php.net/operators.precedence (especially the last paragraph and example). Documentation for the technical reasons behind this can be found at https://gist.github.com/nikic/6699370.
 
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