[PHP-DEV] Bug #10668 Updated: preg_replace backquote failure From: andrei <email protected>
Date: 05/04/01

ID: 10668
Updated by: andrei
Reported By: smoonen <email protected>
Status: Analyzed
Bug Type: PCRE related
PHP Version: 4.0.5
Assigned To:
Comments:

Oops, \\1 instead of \1.

Previous Comments:
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[2001-05-04 15:17:15] andrei <email protected>
Ok, I made a little mistake, use "f('\\1')" rather than "f('1')".

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[2001-05-04 13:59:27] smoonen <email protected>
Oddly, I get the following error:

  Warning: Unexpected character in input: '' (ASCII=92) state=1 in /home/groups/t/ta/tavi/htdocs/playground/test.php(4) : regexp code on line 1

  Parse error: parse error in /home/groups/t/ta/tavi/htdocs/playground/test.php(4) : regexp code on line 1

  Fatal error: Failed evaluating code: f(''\\''') in /home/groups/t/ta/tavi/htdocs/playground/test.php on line 4

PHP compile options may be seen at:

  http://tavi.sourceforge.net/playground/phpinfo.php

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[2001-05-04 12:58:12] andrei <email protected>
Can you explain why you think it fails? The following sample (slightly modified from yours to dump the $s parameter to function):

$str = "abc'\\''def";
function f($s) { var_dump($s); return "x"; }
print preg_replace("/c(.*)d/e", "f('\1')", $str);

outputs:

string(5) "'\''"
abxef

As it should.

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[2001-05-04 11:44:19] smoonen <email protected>
The following code succeeds on PHP 4.03 and PHP 4.04pl1, but fails on PHP 4.05:

  $str = "abc'\\''def";
  function f($s) { return "x"; }
  print preg_replace("/c(.*)d/e", "f('\1')", $str, -1);

This seems to expose *two* underlying bugs:

  1) There appears to be some problem in the regex state
     machine
  2) There is a definite problem with the replacement of
     the backreference with its corresponding string.

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