On Fri, Oct 11, 2019, 6:47 PM Larry Garfield <la...@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) { ... } > > --Larry Garfield > > -- > PHP Internals - PHP Runtime Development Mailing List > To unsubscribe, visit: http://www.php.net/unsub.php As a warning I think, seems I've encountered those errors many times ago. The solution Nikita propose is good, it doesn't deal with too much. These errors are there already, only just giving it a properly meaningful description. +1 for the proposed description.
