Re: ifsetor like expression in php6

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 >