Re: Tidying code for PHP5.0.5/PHP4.4.0

Jochem Maas wrote:

Jasper Bryant-Greene wrote:

From the PHP manual [2]: | the following examples of passing by reference are invalid: | | foo(bar()); // Produces fatal error since PHP 5.1.0 | foo($a = 5); // Expression, not variable

if foo() expects one args by reference then why is doing: foo($a = 5); ..wrong? I always thought that the expression (in this form) 'returns' the variable? or does it need to be:?

Yes, but how can you modify "$a = 5" from within foo()? It's an expression, not something that can be modified. This one is, admittedly, kinda shaky.

I can understand that the following are wrong: foo(5); // not sure this is wrong.

This is definitely wrong. For example: <?php function foo(&$a) { $a = 6; } foo(5); ?> Is obviously impossible, as it tries to assign the value 6 to the constant value 5. -- Jasper Bryant-Greene Freelance web developer http://jasper.bryant-greene.name/