Jasper Bryant-Greene wrote:
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,
by definition the expression is evaluated _before_ the function is
called - so the expression is not passed to the function, the result
of the expression is passed ... I was under the impression that the the
expression evaluates to a 'pointer' (I'm sure thats bad terminology) to
$a ... which can taken by reference by the function.
possibly I am completely misunderstanding what goes on here.
kinda shaky.
I can understand that the following are wrong:
foo(5); // not sure this is wrong.
This is definitely wrong. For example:
your example make it clear, thanks!
<?php
function foo(&$a) {
$a = 6;
}
foo(5);
?>
Is obviously impossible, as it tries to assign the value 6 to the
constant value 5.
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