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Reply-To: bishop@php.net
Date: Fri, 11 Oct 2019 14:46:41 -0400
Message-ID: <CAEYWF=6CxLfzQQnVhMtjMPpTNHeLepONd=TmsHBe8XA62Teu-Q@mail.gmail.com>
To: Larry Garfield <larry@garfieldtech.com>
Cc: php internals <internals@lists.php.net>
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Subject: Re: [PHP-DEV] Warning for "confusable" types
From: bishop@php.net (Bishop Bettini)

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On Fri, Oct 11, 2019 at 1:47 PM Larry Garfield <larry@garfieldtech.com>
wrote:

> On Fri, Oct 11, 2019, at 8:54 AM, Nikita Popov wrote:
> > Hi internals,
> >
> > Something I've seen play out a couple of times: Newbies try to use
> > something like "integer" or "resource" as a type, and then get a
> confusing
> > error message along the lines of "must be an instance of resource,
> resource
> > given".
> >
> > I would like to throw a compiler warning in this case, that looks as one
> of
> > the following:
> >
> > > Warning: "integer" will be interpreted as a class type. Did you mean
> > "int"? Use qualified name or "use" to suppress this warning
> >
> > > Warning: "resource" is not a supported builtin type and will be
> > interpreted as a class type. Use qualified name or "use" to suppress this
> > warning
> >
> > This warning only triggers if the type is lowercase (integer but not
> > Integer), is unqualified (integer but not \integer) and is not imported
> > (there is no "use integer"). This provides multiple ways to avoid the
> > warning for code that does legitimately want to use lowercase "integer"
> as
> > a class type.
> >
> > Implementation: https://github.com/php/php-src/pull/4815
> >
> > Thoughts?
> >
> > Nikita
>
> Can you clarify where exactly "compiler warning" would be displayed?  Is
> that an E_WARNING?  How would I as a user see a message when I write
>
> function foo(integer $a) { ... }
>

It's an E_COMPILE_WARNING, and arises when the compiler sees you're type
hinting for a class, but no such class loaded. This checks that given class
against a known list of "probable mistakes" and provides guidance. Eg, I
typed "integer $a" when I mean "int $a", and don't understand the current
compiler warning: "Argument 1 passed to foo must be an instance of integer,
int given"

See also https://stackoverflow.com/a/40638544/2908724

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