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Date: Sat, 26 Jan 2008 12:22:09 -0500
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Subject: Re: [PHP-DEV] ifsetor like expression in php6
From: sam@sambarrow.com (Sam Barrow)

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
>