[SR-6554] Type inference error when applying optional function to optional value

[swift-ci]

Previous ID	SR-6554
Radar	None
Original Reporter	trimm (JIRA User)
Type	Bug
Status	Resolved
Resolution	Done

Environment

Swift 4.0, Swift 4.0.1

Additional Detail from JIRA

Votes	0
Component/s	Compiler
Labels	Bug, 4.0Regression
Assignee	None
Priority	Medium

md5: d5270fb56c4a13ba96b13a49017fdb53

Issue Description:

public extension Optional {
    public func apply<Result>(_ transform: ((Wrapped) -> Result)?) -> Result? {
        return self.flatMap { value in
            transform.map { $0(value) }
        }
    }

    public func apply<Value, Result>(_ value: Value?) -> Result?
        where Wrapped == (Value) -> Result
    {
        return value.apply(self)
    }
}

Possibly related to bug: https://bugs.swift.org/browse/SR-6509

This works fine under Swift3-3.2.

Actual results:

'apply' produces 'Result?', not the expected contextual result type 'Result?'

Expected results:

The code compiles, compiler infers the type of result for type of Wrapped

[belkadan]

You're supposed to put the second function in a constrained extension. We don't support constraints on generic parameters in an individual function's where clause.

cc @slavapestov

[belkadan]

I guess I should put "supposed" in quotes because clearly it used to work, and it seems to again on master. But I didn't think we supported this.

[slavapestov]

There's no way to express what the user wants with a constrained extension, but the construct is supported. The bug is in the expression checker and not in generics.

[slavapestov]

This is related to function parameter list matching behavior which changed in SE-0110. Looks like something got fixed on master along the way. I'll add a test case.

[slavapestov]

Here is the test: #13345

@belkadan I guess this feature is supported now ;-)

[belkadan]

Oh, I misread the example. I guess this does make sense.