为什么这有效？ Swift（UnsafeMutablePointer <optional>）

Question

protocol Property {}
protocol OptionalProtpery : Property {
        static func codeNilInto(pointer: UnsafePointer<Int>)
    }

    extension Optional : OptionalProtpery {
        static func codeNilInto(pointer: UnsafePointer<Int>) {
            (UnsafeMutablePointer(pointer) as UnsafeMutablePointer<Optional>).memory = nil
        }
    }

我对此声明UnsafeMutablePointer<"Optional">感兴趣 例如，为什么我们没有错误：＆＃34;可选的泛型类型并且需要参数＆lt; ...＆gt;＆＃34;。由于此处未指定特定类型（例如UnsafeMutablePointer＆lt;＆＃34; Optional＆＃34;＆lt;＆＃34; Int＆＃34;＆gt;＆gt;＆gt;）。

Answer 1

protocol Property {}
protocol OptionalProtpery : Property {
    static func codeNilInto(pointer: UnsafePointer<Int>)
}

extension Optional : OptionalProtpery {
    static func codeNilInto(pointer: UnsafePointer<Int>) {
        (UnsafeMutablePointer(pointer) as UnsafeMutablePointer<Optional>).memory = nil
    }
}

var i = 1
let p = withUnsafePointer(&i) { (p) -> UnsafePointer<Int> in
    return p
}
print(p.memory) // 1

//Optional.codeNilInto(p) // error: generic parameter 'Wrapped' could not be inferred
// you have to specify generic parameter !!!
Optional<Int>.codeNilInto(p)
Optional<Double>.codeNilIntro(p)
// or what ever you want to express

所以，因为Optional是通用的，所以你的代码是'OK'。顺便说一下，您对代码的期望是什么？