Scala:如何将可变参数指定为类型? [英] Scala: how to specify varargs as type?

问题描述

代替

def foo(configuration: (String, String)*)

我希望能够写:

type Configuration =  (String, String)*
def foo(configuration: Configuration)

主要用例是在子类中覆盖时提供简单的方法签名

The main use case is to provide an easy method signature when overriding in subclasses

更新:我可以靠近

type Param = (String, String)
def foo(configuration: Param*)

但是有没有更好的方法?

But is there a way of doing it better?

推荐答案

不，* 只允许用于 ParamType，即匿名函数或方法的参数类型.

No, the * is only allowed on a ParamType, that is the type of a parameter to a anonymous function or a method.

4.6.2 重复参数语法:ParamType ::= Type ‘’ 最后一个值参数部分的参数可以以"为后缀，例如(..., x:T *).这种重复参数的类型方法内部是序列输入 scala.Seq[T].方法与重复参数 T * 取一个可变数量的 T 类型参数.

4.6.2 Repeated Parameters Syntax: ParamType ::= Type ‘’ The last value parameter of a parameter section may be suffixed by "", e.g. (..., x:T *). The type of such a repeated parameter inside the method is then the sequence type scala.Seq[T ]. Methods with repeated parameters T * take a variable number of arguments of type T.

编译器错误@Eastsun 的例子在第一行，而不是第二行.这不应该被允许:

The compiler bug @Eastsun's example is in the first line, not the second. This should not be allowed:

scala> type VarArgs =  (Any*)
defined type alias VarArgs

我提出了一个错误.

这与按名称参数的限制类似.在这种情况下，编译器会阻止创建类型别名:

This is similar restriction to By-Name Parameters. In this case, the compiler prevents the creation of the type alias:

scala> type LazyString = (=> String) <console>:1: error: no by-name parameter type allowed here
       type LazyString = (=> String)

你最后的尝试是表达这一点的标准方式.

Your final attempt is the standard way to express this.

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