ifsetor like expression in php6 View as Markdown

Hi,

Am Samstag, den 26.01.2008, 12:17 -0500 schrieb Sam Barrow:

I don't think throwing a E_NOTICE is appropriate. The isset() construct
doesn't throw an E_NOTICE, this shouldn't either.

As far as I understand it is just an extension to the already present
tertiary operator and therefore the ifsetor would be
isset($_GET['foo']) ?: 'bar'. I don't understand why that sugar should
act anything different and not trigger an E_NOTICE.

cu, Lars

»Aber das Verhältnis von Leben und Produktion, das jenes
real herabsetzt zur ephemeren Erscheinung von dieser, ist
vollendet widersinnig. Mittel und Zweck werden vertauscht.«
-- Theodor W. Adorno, »Minima Moralia«, Die traurige Wissenschaft

Lars Strojny
Nießenstr. 36
51103 Cologne
Jabber/Mail: lars@strojny.net
Weblog: http://usrportage.de

On Saturday 26 January 2008 18:22:09 Sam Barrow wrote:

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 ;
}

That magically brings $foo into life if you do ifsetor($foo, 2).

Hi,
usage of references here have some bad consequences, it will pollute
the input array with null entries:

function foo(&$a) { }
var_dump(array_key_exists('inexistent', $_GET)); // bool(false)
foo($_GET['inexistent']);
var_dump(array_key_exists('inexistent', $_GET)); // bool(true)

will actually create the inexistent entry and fill it with null.
Again, there is no perfect implementation of an issetor in userland.

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.

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

--

--
Etienne Kneuss
http://www.colder.ch

Men never do evil so completely and cheerfully as
when they do it from a religious conviction.
-- Pascal

Hello Sam,

as others pointed out & results in unwanted entries being present. So the
second time you use ifsetor on those entries, you do not get the default
back but rather NULL. Actually in your implementation you never get the
default back.

Please read the archives, all of this has been discussed several times and
we have had a bunch of in depth explanations. In fact our archives hold a
lot of information valid to people new to the list as well as for us
following the list for years. I guess I am not the only one having my own
searchable archive always with me - for a reason.

marcus

Saturday, January 26, 2008, 6:22:09 PM, you wrote:

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.

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

Best regards,
Marcus