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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 00:55:10 +0100
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Cc: Stanislav Malyshev <stas@zend.com>
References: <200801031903.01980.tomi@cumulo.fi> <200801040008.43944.pstradomski@gmail.com> <477D6E68.5090501@zend.com>
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Message-ID: <200801040055.10360.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 pi=C4=85tek 04 stycznia 2008:
> > 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) wou=
ld
> > be
>
> Why would you want it? I wouldn't want my code to die, I would want it
> to work.

You wrote:

> With strict checking, that means instead of calling:
> foo($bar);
> you'd have now to do:
> if(is_integer($bar)) {
> =C2=A0 =C2=A0 // stupid foo() call would blow up if passed not integer, s=
o I have
> to manually handle it
> =C2=A0 =C2=A0 // oh wait, actually I don't have any way to handle it - I =
need
> foo() to be called to proceed
> =C2=A0 =C2=A0 // so let's just
> =C2=A0 =C2=A0 die("can't call foo() because \$bar is not integer");
> }

This shows that you explicitly die() if $bar is not integer. But with stric=
t=20
type hinting this is unneccessary - the script would automatically die if=20
$bar was not integer. In other words, with strict scalar type hinting the=20
above would behave just like=20

foo($bar)

with the only difference being the error message (in your version you've go=
t=20
your own error message, automatic script failure would raise some built-in=
=20
error - which IMO should be E_RECOVERABLE_ERROR, as E_FATAL gives blano=20
output on production sites where display_error is 0).

Not that I care if scalar type hinting is implemented or not, I just wanted=
 to=20
point out that your example is just plain wrong.

>
> > you have either an int or string representing an integer in $bar, then
> > foo((integer)$bar) would work just well.
>
> And then how exactly strict typing helps you?

Here it doesn't help, but is easy to work around if you are sure you know w=
hat=20
you're doing. This is the same workaround that is needed if someone used=20

if (! is_integer($bar)) {die()}

inside of the foo method.

=2D-=20
Pawe=C5=82 Stradomski