正如标题中所述:为什么可以忽略子类型的重写成员函数中的默认值？
Is this normal or to be expected?
interface Foo {
fun bar(parameter: Int = 1)
}
class Baz : Foo {
override fun bar(parameter: Int) { // OK
println(parameter)
}
}
val baz = Baz()
baz.bar() // OK
baz.bar(2) // OK
Same behavior in the case where Foo is a class.
这是正常的还是意料之中的？
对
I suspect this was primarily a language design / usability decision. From this perspective, there were four options available:
The Kotlin designers chose option #1. This makes sense because:
选项#2和#4都暗示调用者不知道默认值是什么，除非他们知道自己使用的是哪个实现，这当然是非常不可取的.调用方需要额外的逻辑来确定是否需要一个值来覆盖默认值，这意味着默认值将是无用的.
选项3违反了DRY principle.为什么要把声明放在两个地方？
That leaves Option #1 as the only sane choice.