Kotlin：如何调用扩展方法的超级实现

Question

在 Kotlin 中，我有如下代码：

open class A {
    protected open fun B.doSomething() {}
}

class B {}

class C : A() {

    override fun B.doSomething() {
        print("first, do some C-specific logic")
        print("now try to call super implementation. How?")
        super.doSomething() // does not compile
        super<A>.doSomething() // does not compile
        super<B>.doSomething() // does not compile
    }
}

而且我想知道，在这种情况下如何调用 doSomething() 的超类实现？或者根本不可能在 kotlin 中为扩展函数调用超级实现？

Answer 1

此问题已在 KT-11488 中报告。目前不可能。 Kotlin 规范说 super-form 调用的候选成员是通过以下方式找到的：

For a callable f with an explicit basic super-form receiver super in a classifier declaration with supertypes A1, A2, … , AN the following sets are considered for non-emptiness:

Non-extension member callables named f of type A1;

Non-extension member callables named f of type A2;

…;

Non-extension member callables named f of type AN.

If at least two of these sets are non-empty, this is a compile-time error. Otherwise, the non-empty set (if any) is analyzed as usual.

For a callable f with an explicit extended super-form receiver super<A> the following sets are analyzed (in the given order):

Non-extension member callables named f of type A.

如您所见，仅考虑“非扩展成员可调用文件”。完全不考虑扩展功能。

现在，您可以像这样解决（票证中也建议类似的方法）：

open class A {
    protected open fun B.doSomething() =
        bDoSomething(this)

    // make a non-extension function to put A's implementation in there
    protected fun bDoSomething(receiver: B) {
        print("A implementation")
    }
}

class C : A() {

    override fun B.doSomething() {
        print("first, do some C-specific logic")
        super.bDoSomething(this) // call the non-extension function
    }
}

Kotlin：如何调用扩展方法的超级实现

Kotlin: How to call super implementation of an extension method

extension-methods

overriding

kotlin