是否有可能将Swift泛型类的函数返回类型限制为相同的类或子类？ [英] Is it possible to restrict a Swift generic class function return type to the same class or subclass?

问题描述

我在Swift中扩展了一个基类（我不控制它）。我想提供一个用于创建类型为子类的实例的类函数。通用函数是必需的。但是，类似下面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()

调试器报告 T as ScientificCalculator ，但 sci 是计算器并调用 sci.showKind（）返回regular。

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

有没有一种方法可以使用泛型来实现所需的结果，还是一个bug？

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

推荐答案

好的，从开发人员论坛，如果您拥有基类的控制权，

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()

不幸的是，如果你没有控制基类，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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