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Reply-To: =?utf-8?q?Pawe=C5=82_Stradomski?= <pstradomski@gmail.com>
To: internals@lists.php.net
Date: Fri, 4 Jan 2008 00:08:43 +0100
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Cc: Stanislav Malyshev <stas@zend.com>
References: <200801031903.01980.tomi@cumulo.fi> <cd5680a20801031412p387103d5k694eed0f2978eeda@mail.gmail.com> <477D66B4.2000107@zend.com>
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Message-ID: <200801040008.43944.pstradomski@gmail.com>
Subject: Re: [PHP-DEV] RE: Optional scalar type hinting
From: pstradomski@gmail.com (=?utf-8?q?Pawe=C5=82_Stradomski?=)

W li=C5=9Bcie Stanislav Malyshev z dnia czwartek 03 stycznia 2008:
> > type-hinting is asserting.
> > checking of types is needed only on interface-border points (where
>
> With strict checking, that means instead of calling:
> foo($bar);
> you'd have now to do:
> if(is_integer($bar)) {
>     // stupid foo() call would blow up if passed not integer, so I have
> to manually handle it
>     // oh wait, actually I don't have any way to handle it - I need
> foo() to be called to proceed
>     // so let's just
>     die("can't call foo() because \$bar is not integer");
> }
> foo($bar);
>

This code is just a good argument *FOR* type hints. When foo is:
function foo(int $bar) {...}
and you want the code to die if $bar is not integer, then foo($bar) would b=
e=20
the way to go, without the need for the mess you posted. And if you know th=
at=20
you have either an int or string representing an integer in $bar, then=20
foo((integer)$bar) would work just well.

The only problem with the cast is that (integer)'abc' is 0, but it has been=
 so=20
for a long time and there is no need to change it. is_numeric check solves=
=20
the problem.


=2D-=20
Pawe=C5=82 Stradomski