candidate 将匹配并推断 'T' = '[U]' 如果 '[U]' 符合 'Foo'

Question

如果这样的代码，将显示错误并且编译器无法推断 T = [U]。

protocol Foo { }

protocol Bar {
    associatedtype T: Foo
    var type: T { get }
}

struct Baz<U: Foo> {
    let type: [U]
}

extension Baz: Bar { }

/* error message
error: Demo.playground:14:1: error: type 'Baz<U>' does not conform to protocol 'Bar'
extension Baz: Bar { }
^

Demo.playground:6:20: note: unable to infer associated type 'T' for protocol 'Bar'
    associatedtype T: Foo
                   ^

Demo.playground:11:9: note: candidate would match and infer 'T' = '[U]' if '[U]' conformed to 'Foo'
    let type: [U]
        ^
*/

但是将Foo替换为Decodable，编译器可以推断出T = [U]，为什么？

// protocol Foo { }

protocol Bar {
    associatedtype T: Decodable
    var type: T { get }
}

struct Baz<U: Decodable> {
    let type: [U]
}

extension Baz: Bar { }

Answer 1

Array 有此扩展名：

extension Array: Encodable where Element: Encodable …

照样做。

extension Array: Foo where Element: Foo { }

candidate 将匹配并推断 'T' = '[U]' 如果 '[U]' 符合 'Foo'

candidate would match and infer 'T' = '[U]' if '[U]' conformed to 'Foo'

generics

protocols

swift