You have added is_int, is_numeric, and === operator. I've been reading your
posts and you should really stop confusing using that argument. Truth is,
types are relevant.
$foo = 1;
$bar = '1';
if($foo === $bar) {
echo "I am never echo'ed";
}
Until you make $foo === $bar, you should really stop telling everyone how
strict param/return types have no place in php.
-Chris
On Wed, Apr 30, 2008 at 9:36 AM, Stanislav Malyshev <[EMAIL PROTECTED]> wrote:
> Hi!
>
> Right, return type hinting is useless in such cases, but it is
> > interesting when we check whether a method is compatible with an
> > implemented interface or an extended/abstract class. In this case,
> >
>
> When exactly you check that? PHP has no static type checking, and in
> runtime actual values, not function definitions, are checked. If you mean
> checking by tools like IDE, phpdoc is more than enough for that.
> --
> Stanislav Malyshev, Zend Software Architect
> [EMAIL PROTECTED] http://www.zend.com/
> (408)253-8829 MSN: [EMAIL PROTECTED]
>
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