重载字典下标两次并转发调用

问题描述

我正在尝试扩展 Dictionary 并允许提取转换为特定类型和给定默认值的值.为此，我为 subscript 函数添加了两个重载，一个有默认值，一个没有:

I'm trying to extend Dictionary and allow extracting values casted to a certain types and with a given default value. For this I added two overloads for the subscript function, one with a default value, one without:

extension Dictionary {

    subscript<T>(_ key: Key, as type: T.Type, defaultValue: T?) -> T? {
        // the actual function is more complex than this :)
        return nil
    }

    subscript<T>(_ key: Key, as type: T.Type) -> T? {
        // the following line errors out:
        // Extraneous argument label 'defaultValue:' in subscript
        return self[key, as: type, defaultValue: nil]
    }
}

但是，当从二元下标调用三元下标时，出现以下错误:

However when calling the three-argument subscript from the two-argument one I get the following error:

下标中的无关参数标签defaultValue:"

这是 Swift 的限制吗?还是我遗漏了什么?

Is this a Swift limitation? Or am I missing something?

我使用的是 Xcode 10.2 beta 2.

I'm using Xcode 10.2 beta 2.

附言我知道还有其他替代方法，例如专用函数或 nil 合并，试图了解在这种特定情况下出了什么问题.

P.S. I know there are other alternatives to this, like dedicated functions or nil coalescing, trying to understand what went wrong in this particular situation.

推荐答案

当涉及到参数标签时，下标与函数有不同的规则.对于函数，参数标签默认为参数名称——例如，如果您定义:

Subscripts have different rules than functions when it comes to argument labels. With functions, argument labels default to the parameter name – for example if you define:

func foo(x: Int) {}

您可以将其称为 foo(x: 0).

然而对于下标，参数默认没有参数标签.因此，如果您定义:

However for subscripts, parameters don't have argument labels by default. Therefore if you define:

subscript(x: Int) -> X { ... }

你会称它为 foo[0] 而不是 foo[x: 0].

you would call it as foo[0] rather than foo[x: 0].

因此在您的示例中带有下标:

Therefore in your example with the subscript:

subscript<T>(_ key: Key, as type: T.Type, defaultValue: T?) -> T? {
    // the actual function is more complex than this :)
    return nil
}

defaultValue: 参数没有参数标签，这意味着下标必须被称为 self[key, as: type, nil].为了添加参数标签，您需要指定两次:

The defaultValue: parameter has no argument label, meaning that the subscript would have to be called as self[key, as: type, nil]. In order to add the argument label, you need to specify it twice:

subscript<T>(key: Key, as type: T.Type, defaultValue defaultValue: T?) -> T? {
    // the actual function is more complex than this :)
    return nil
}

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