'Generic parameter 'T' could not be inferred'

Question

I don't know why this happening, I'd like to return a parameter in accordance to MyProtocol with an array, so I'm trying to achieve that the following way:

import Foundation

protocol Test {

}

protocol ArrayTest {
    associatedtype E : Test
}

extension Array : ArrayTest where Element:Test{
    typealias E = Element
}

class BBB {
 func ggg<T:ArrayTest>( kk: ((_ value:T) -> Void)?){

 }
}

class AAA<T:ArrayTest> {

    func ggg(){
     BBB().ggg { (test) in//Generic parameter 'T' could not be inferred 
 here

     }
 }
 }

What type do you want `test` to be in the line `BBB().ggg { (test) in`? — Sweeper, Oct 18 '18 at 06:52
Sometimes it still seems the compiler is not as clever as we are, so you need to explicitly annotate the type: `BBB().ggg { (test:T) in // ...` — Andreas Oetjen, Oct 18 '18 at 06:55
Better naming (using "real world" names, not `ggg` and `kk`, etc ), and explaining what you're trying to achieve would help. — Ashley Mills, Oct 18 '18 at 07:02

score 2 · Accepted Answer · answered Oct 18 '18 at 07:08

You can't just do BBB().ggg { (test: [Test]) in ... because [Test] does not conform to ArrayTest.

This might come as surprising to you, as you wrote:

extension Array : ArrayTest where Element: Test{
    typealias E = Element
}

Surely the above extension applies to [Test], right? No. For the above extension to apply, Element must conform to Test, but Test does not conform to itself.

I think what you might mean is this:

class AAA<T:ArrayTest> {

    func ggg(){
        BBB().ggg { (test: T) in

        }
    }
}

This uses T from the declaration of AAA.

And as the user of the AAA class, you need to pass an actual type that conforms to ArrayTest, such as [U], where U is a type conforming to Test.

It works!,Thank you very much.this help me a lot,But i still need to understand for a while — Dandelion, Oct 18 '18 at 07:19
@Dandelion please consider accepting my answer if you find this helpful. — Sweeper, Oct 18 '18 at 07:20

Ilya Kharabet · Answer 2 · 2018-10-18T07:11:43.107

0

The compiler can't define a type of generic parameter T. You need to specify the type of T explicitly:

BBB()ggg { (test: [Test]) in

}

edited Oct 18 '18 at 07:11

answered Oct 18 '18 at 07:03

Ilya Kharabet

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'Generic parameter 'T' could not be inferred'

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