note 99546 modified in language.references.pass by danbrown

From: Date: Tue, 24 Aug 2010 03:39:48 +0000
Subject: note 99546 modified in language.references.pass by danbrown
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Passing variable reference to function instead of declaring the function parameter type to reference also could get the result as expected. <?php $string = 'string'; function change($str) { $str = 'str'; } change(&$string); echo $string; ?> yeild str --was-- Passing variable reference to function instead of declaring the function parameter type to reference also could get the result as expected. $string = 'string'; function change($str) { $str = 'str'; } change(&$string); echo $string; yeild str http://php.net/manual/en/language.references.pass.php

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