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Bug #30732 Wrong loop handling when using chars instead of integers
Submitted: 2004-11-09 11:07 UTC Modified: 2004-11-09 16:44 UTC
From: thomas dot keller at inatec dot com Assigned:
Status: Not a bug Package: Scripting Engine problem
PHP Version: 4.3.9 OS: Linux Mandrake 8, Linux 2.4.3
Private report: No CVE-ID: None
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 [2004-11-09 11:07 UTC] thomas dot keller at inatec dot com 
Description:
------------
Configure command:
'./configure' '--prefix=/usr/local/php' '--with-apxs2=/home/apache2/apache-2.0.52/bin/apxs' '--enable-versioning' '--with-mysql=/usr/local/mysql' '--with-gd' '--with-zlib'

I use the following code snippet to print out an array
of chars in alphabetic order. If I have the condition
$i < 'Z' (so only until 'Y'), everything works fine, 
for the condition(s) $i <= 'Z' or $i < ('A' + 26)
PHP produces wrong results.

I could reproduce the error on a Redhat 9 system with PHP 4.3.4.

Reproduce code:
---------------
$letters = array();

for ($i='A'; $i<='Z'; $i++) print $i;



Expected result:
----------------
ABCDEFGHIJKLMNOPQRSTUVWXYZ

(26 ^ 1 letters)

Actual result:
--------------
ABCDEFGHIJKLMNOPQRSTUVWXYZAAABAC .. YZ

(26 ^ 2 letters)

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 [2004-11-09 15:09 UTC] mgf@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

When you use ++ on strings, the sequence of values is:

... \'X\', \'Y\', \'Z\', \'AA\', \'AB\', ... \'AZ\', \'BA\', ...

Only when this sequence reaches \'ZA\' does it fail the <=\'Z\' test.
 [2004-11-09 16:44 UTC] thomas dot keller at inatec dot com 
Hrm... I thought of 'Z' not as a string, but
as a single character. In other programming
languages (C/C++, Java) a char is a signed
BYTE, thus can be used as replacement for
integer values. So I expect I need to write

for ($i=ord('A');$i<=ord('Z'); $i++) print chr(i);

to get my desired output? Since I don't know
a PHP shorthand for Perl's

foreach ('A' .. 'Z') print $_;

Thanks for clearing that up.
 
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