param: _* 在 Scala 中是什么意思? [英] What does param: _* mean in Scala?

问题描述

作为 Scala (2.9.1) 的新手，我有一个 List[Event] 并且想将它复制到一个 Queue[Event]，但以下内容语法产生一个 Queue[List[Event]] 代替:

Being new to Scala (2.9.1), I have a List[Event] and would like to copy it into a Queue[Event], but the following Syntax yields a Queue[List[Event]] instead:

val eventQueue = Queue(events)

出于某种原因，以下有效:

For some reason, the following works:

val eventQueue = Queue(events : _*)

但我想了解它的作用，以及它为什么起作用?我已经看过 Queue.apply 函数的签名:

But I would like to understand what it does, and why it works? I already looked at the signature of the Queue.apply function:

def apply[A](elems: A*)

我明白为什么第一次尝试不起作用，但第二次尝试是什么意思?什么是 : 和 _* 在这种情况下，为什么 apply 函数不只采用 Iterable[A] ?

And I understand why the first attempt doesn't work, but what's the meaning of the second one? What is :, and _* in this case, and why doesn't the apply function just take an Iterable[A] ?

推荐答案

a: A 是类型归属；参见 Scala 中类型归属的目的是什么?

: _* 是类型归属的特殊实例，它告诉编译器将序列类型的单个参数视为可变参数序列，即 varargs.

: _* is a special instance of type ascription which tells the compiler to treat a single argument of a sequence type as a variable argument sequence, i.e. varargs.

使用 Queue.apply 创建一个 Queue 是完全有效的，该Queue.apply 具有单个元素是序列或可迭代的，所以这正是当你给单个 Iterable[A].

It is completely valid to create a Queue using Queue.apply that has a single element which is a sequence or iterable, so this is exactly what happens when you give a single Iterable[A].

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