I'm sorry I misunderstood. If specifying like "$var ?: 5" then it should
throw an E_NOTICE, as this is a conditional that checks the value of a
variable ($var).
Sebastian, for assigning of a default value is a variable is not set, I
would recommend using something like this (this is what I use):
function ifsetor(&$var, $or) {
return isset($var) ? $var : $or ;
}
Because $var is passed by reference, it won't throw an E_NOTICE when
passing an undeclared variable.
On Sat, 2008-01-26 at 17:30 +0100, Sebastian wrote:
> hi,
>
> i just found the new ifsetor like construction in php6. It left me the
> question why php has to throw an E_NOTICE when the variable is not set. This
> is basicly because that actually is the meaning of this construction,
> checking if the variable is set and if it isnt give back a standard value.
> Of course i know this construction can be (mis)used with an Boolean check
> ($var===$var2), but this does not make much sence either because TRUE will
> then be assigned to the new variable if the expression evaluates to TRUE
> (but please correct me if theres a real use for this).
>
> So in any case it would be better to remove the Boolean check feature and
> make it a simple ifsetor() without the E_NOTICE again.
>
> For those of you who have not heard of it heres an example:
>
> <?php
>
> // If $_GET['foo'] is set, then its value will be assigned to $foo,
> // otherwise 42 will be assigned to $foo.
>
> $foo = $_GET['foo'] ?: 42;
>
> ?>
>
> Greetings
>
> Sebastian
>
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