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Subject: Re: [PHP-DEV] [RFC] Default expression
In-Reply-To: <B2BAE1A6-D2A1-49F6-85A0-CD83E07ACB75@rwec.co.uk>
Date: Tue, 27 Aug 2024 11:03:40 -0400
Cc: internals@lists.php.net,
 Bilge <bilge@scriptfusion.com>
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Message-ID: <A8CD76C6-7409-4C78-BEFD-6782059FD9B1@zort.net>
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To: "Rowan Tommins [IMSoP]" <imsop.php@rwec.co.uk>
From: jbafford@zort.net (John Bafford)


> On Aug 27, 2024, at 10:14, Rowan Tommins [IMSoP] =
<imsop.php@rwec.co.uk> wrote:
>=20
> The only expressions that are in some sense "safe" are those that can =
apply equally to any possible type that the function could in future set =
as the default. In theory, that includes a match statement with an arm =
of "default =3D> default", e.g.
>=20
> json_encode($data, default, match(gettype(default)) { 'int' =3D> =
default | JSON_PRETTY_PRINT, default =3D> default });
>=20
> Apart from being incredibly hard to read, that's not even useful: the =
aim is to always enable pretty printing, but the result is "enable =
pretty print, unless the type of the default happens to change".


I'm not sure this could even work at all. The "default" parameter to =
gettype() isn't the default value of the third parameter to =
json_encode(). It's the default value of the first parameter to =
gettype(). Which would probably fail, since gettype()'s first parameter =
doesn't have a default. I suppose this could be solved by specifying an =
offset or label (e.g. as with `continue 2` in a nested loop), but that =
would just make it even harder to read.

-John