为什么Kotlin无法覆盖List< *>运算符方法? [英] Why Kotlin can not override List<*> operator method?

问题描述

这是我的职能:

operator  infix fun  List<Teacher>.get(int: Int): Teacher {
    var t = Teacher()
    t.name = "asd"
    return t ;
}

和我的用法:

b[0].teachers[1].name

提示:b是具有List<老师>财产

tip: b is an object that has List< Teacher > property

和错误Empty list doesn't contain element at index 1.

为什么此替代运算符功能不起作用?

why this override operator function doesn't work?

推荐答案

在Kotlin中，您不能隐藏带有扩展名的成员函数. 成员始终会在通话解决方案中获胜.因此，您基本上不能调用具有与成员函数相同的签名的扩展名，该扩展名存在于为表达式声明或推断的类型中.

In Kotlin, you cannot shadow a member function with an extension. A member always wins in the call resolution. So, you basically cannot call an extension with a signature same to that of a member function, that is present in the type that was declared or inferred for the expression.

class C {
    fun foo() { println("member") }
}

fun C.foo() { println("extension") }

C().foo() // prints "member"

在您的情况下，成员函数为在kotlin.collections.List中定义的abstract operator fun get(index: Int): E .

In your case, the member function is abstract operator fun get(index: Int): E defined in kotlin.collections.List.

请参见语言参考:扩展名已解析静态

See the language reference: Extensions are resolved statically

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