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) { ... }
--Larry Garfield
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