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Bug #32700 unpack() returns signed values when it shouldn't
Submitted: 2005-04-13 22:49 UTC Modified: 2005-04-14 00:48 UTC
From: ieure@php.net Assigned:
Status: Not a bug Package: Strings related
PHP Version: 4.3.10 OS: Linux
Private report: No CVE-ID: None
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 [2005-04-13 22:49 UTC] ieure@php.net 
Description:
------------
unpack() returns signed long values when the high bit is  
set, even though the unpack format is set to unsigned long.  
  
Workaround:  
  
$foo = unpack('N', $data);  
$foo = $foo << 1 >> 1;  
$foo += 0x80000000; 
   
   

Reproduce code:
---------------
// High bit is set.
$before = 0x80000002;
echo "Before: $before\n"; // Outputs '2147483650'
$int = pack('N', $before);
$res = unpack('N', $int);
$after = $res[1];
echo "After: $after\n"; // Outputs '-2147483646'


Expected result:
----------------
$before and $after should be the same number. 

Actual result:
--------------
$after is treated as signed, and does not equal $before.  
 
Script output: 
 
Before: 2147483650 
After: -2147483646

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 [2005-04-13 22:54 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

PHP only supports signed integers.
 [2005-04-13 22:54 UTC] derick@php.net 
try printf("%u") though to print it - that would work...
 [2005-04-13 23:01 UTC] ieure@php.net 
While PHP may only support signed integers, the manual for 
pack() clearly states, regarding the 'N' format: 
 
"unsigned long (always 32 bit, big endian byte order)" 
 
But it returns a signed int when using this format, 
contrary to the documentation. 
 
There's obviously some sort of internal PHP state regarding 
the signedness of a particular value, since it's capable of 
correctly representing the unsigned value. It just seems 
that unpack() is doing something wrong here.
 [2005-04-13 23:07 UTC] derick@php.net 
Yes, but echo will show the signed representatation. This is *not* a bug, use printf("%u") to show the unsigned version.
 [2005-04-13 23:14 UTC] ieure@php.net 
It has nothing do with "echo showing the signed version." 
If that were the case, it would output the same (negative) 
number when you echo $before as well. 
 
It's PHP /using/ the signed version. For example:    
    
$test[$before] = 'It works!';    
if (isset($test[$after])) {    
    echo "It works!\n";   
} else {   
    echo "It doesn't work!\n";  
}  
  
$before and $after are /not/ the same value to PHP. Nothing 
to do with echo.
 [2005-04-13 23:58 UTC] sniper@php.net 
RTFM.. (for pack() function)
 [2005-04-13 23:59 UTC] derick@php.net 
And it's still not a bug.
 
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