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Date: Fri, 3 Jun 2011 21:40:57 -0700
Message-ID: <BANLkTi=aZERFNhKa4izz+jdZa6efdKA5ng@mail.gmail.com>
To: =?ISO-8859-1?Q?David_Z=FClke?= <david.zuelke@bitextender.com>
Cc: Martin Scotta <martinscotta@gmail.com>, Sean Coates <sean@seancoates.com>, 
	Anthony Ferrara <ircmaxell@gmail.com>, PHP internals <internals@lists.php.net>
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Subject: Re: [PHP-DEV] RFC: Short syntax for Arrays (redux)
From: andrei@zmievski.org (Andrei Zmievski)

On Fri, Jun 3, 2011 at 3:11 PM, David Z=FClke
<david.zuelke@bitextender.com> wrote:
> It's not FUD.
>
> It is different from writing json_decode('=E4\u0123'), because json_decod=
e() in PHP only accepts UTF-8 encoded input;
>
> Give it a shot:
>
> <?php
> $chr =3D "\xC3\xA4"; // "=E4" as UTF-8
> var_dump(json_decode('["' . $chr . '\u00e4"]'));
> var_dump(json_decode('["' . utf8_decode($chr) . '\u00e4"]'));
> ?>
>
> That'll produce:
>
>> array(1) {
>> =A0 [0]=3D>
>> =A0 string(4) "=E4=E4"
>> }
>> NULL
>
> Understand what the problem is now?
>
> If someone does this in a latin1-encoded file:
>
> <?php $fancyNewArray =3D {"yay": "=E4"}; ?>
>
> Then that is valid as a PHP array (as it's a latin1 "=E4", so \xE4), but =
cannot be consumed by PHP's json_decode(). And that would be terribly incon=
sistent behavior.
>
> David

Of course I know how json_decode() works, David. My question back to
you is, do you know how a parser works? Because there's no reason to
invoke json_decode() on $fancyNewArray, since it would already be an
object, constructed by the Zend parser. The handling of the characters
in the key/value strings is exactly the same as in normal object
property assignments.

-Andrei

P.S. Stop being so patronizing.