问题描述

我正在 Swift 中扩展一个基类(我无法控制).我想提供一个类函数来创建一个子类类型的实例.需要一个通用函数.但是，像下面这样的实现不会返回预期的子类类型.

I am extending a base class (one which I do not control) in Swift. I want to provide a class function for creating an instance typed to a subclass. A generic function is required. However, an implementation like the one below does not return the expected subclass type.

class Calculator {
    func showKind() { println("regular") }
}

class ScientificCalculator: Calculator {
    let model: String = "HP-15C"
    override func showKind() { println("scientific") }
}

extension Calculator {
    class func create<T:Calculator>() -> T {
        let instance = T()
        return instance
    }
}

let sci: ScientificCalculator = ScientificCalculator.create()
sci.showKind()

调试器将 T 报告为 ScientificCalculator，但 sci 是 Calculator 并调用 sci.showKind() 返回常规".

The debugger reports T as ScientificCalculator, but sci is Calculator and calling sci.showKind() returns "regular".

有没有办法使用泛型来达到预期的结果，或者它是一个错误?

Is there a way to achieve the desired result using generics, or is it a bug?

推荐答案

好的，来自开发者论坛，如果您可以控制基类，则可以实现以下解决方法.

Ok, from the developer forums, if you have control of the base class, you might be able to implement the following work around.

class Calculator {
    func showKind() { println("regular") }
    required init() {}
}

class ScientificCalculator: Calculator {
    let model: String = "HP-15C"
    override func showKind() { println("(model) - Scientific") }
    required init() {
        super.init()
    }
}

extension Calculator {
    class func create<T:Calculator>() -> T {
        let klass: T.Type = T.self
        return klass()
    }
}

let sci:ScientificCalculator = ScientificCalculator.create()
sci.showKind()

很遗憾，如果您无法控制基类，则这种方法是不可能的.

Unfortunately if you do not have control of the base class, this approach is not possible.

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