我可以从接口中的另一个泛型声明解析泛型类型吗? [英] Can I resolve a generic type from another generic declaration in an interface?

问题描述

期望您拥有这样的界面:

Expect you have an interface like this:

interface MyInterface<T : BaseClass<I>, I> {
    fun someMethod(param: I) : T
}

如您所见，我在 someMethod 中使用 I 作为参数.但是实际上当我实现这样的接口时，我不想声明 I :

As you can see I use I as a parameter in someMethod. But actually I don't want to declare I when I implement this interface like this:

class BaseClassImpl : BaseClass<OtherClass>

class Impl : MyInterface<BaseClassImpl, OtherClass> {
    override fun someMethod(param: OtherClass) {
        TODO("Not yet implemented")
    }
}

理论上，编译器可以解析 I 泛型而无需附加声明，因为它是由 BaseClassImpl 提供的.因此， MyInterface< BaseClassImpl> 应该已经提供了足够的信息来解决 someMethod()的必要泛型.

Theoretically it should be possible that the I generic can be resolved by the compiler without the additional declaration because it's provided by BaseClassImpl. So MyInterface<BaseClassImpl> should already provide enough information to resolve the necessary generic for someMethod().

在Kotlin中有什么方法可以实现这一目标吗?

Is there any way to achieve that in Kotlin?

推荐答案

在Kotlin中是不可能的.语言规范状态:

It's impossile in Kotlin. Language specification states:

Kotlin支持两种类型推断.

There are two kinds of type inference supported by Kotlin.

• 本地类型推断，用于在语句/表达式范围内本地推断表达式的类型；
• 函数签名类型推断，用于推断函数返回值和/或参数的类型.

它不能基于另一个泛型参数的类型来推断一个泛型参数的类型(尤其是对于超类型声明，因为它是构建类型约束系统的基础).

It can't infer type of one generic parameter based on the type of another (especially for supertype declaration, because it is a very base for building type constrains system).

您可以声明类型别名(针对每个 T )，以避免在每次实现此接口时重复 I :

You may declare typealiases (for each T) to avoid repating I each time you implement this interface:

typealias MyInterfaceForBaseClassImpl = MyInterface<BaseClassImpl, OtherClass>

class Impl : MyInterfaceForBaseClassImpl {
    override fun someMethod(param: OtherClass) : BaseClassImpl {
        //...
    }
}

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