匿名函数上的 scala By-name 参数 [英] scala By-name parameter on a anonymous function

问题描述

我正在努力编写一个带有 by-name 参数的匿名函数.这就是我累了.

I'm struggling to write an anonymous function with by-name parameter. Here is what i tired.

val fun = (x: Boolean, y: =>Int) => if(x) y else 0

此操作失败并出现以下错误.

This fail with following error.

Error:(106, 31) identifier expected but '=>' found.
    val fun = (x: Boolean, y: =>Int) => if(x) y else 0
                              ^
Error:(109, 3) ')' expected but '}' found.
  }
  ^

与标准函数相同的代码如何工作.

How ever same code as a standard function works.

  def fun1(x: Boolean, y: =>Int) = if(x) y else 0

有什么指点吗?

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我遇到了一个由两部分组成的问题.senia 的回答解决了最初的案例.假设我有一个函数需要一个函数.

I had a two part problem. senia answer solved the initial case. Suppose I have a function takes a function.

  def xxx[A,B](f:(A,=>B)=>B)={}

根据 senia 解决方案，它可以工作.

As per senia solution it works.

val fun: (Int, =>Boolean) => Boolean = (x, y) => y
xxx[Int,Boolean](fun)

但是我想摆脱中间的乐趣并使用匿名函数调用xxx.做

However I wanna get rid of the intermediate fun and call xxx with anonymous function. Doing

xxx((Int, =>Boolean) => Boolean = (x, y) => y) 

不会工作.任何想法如何做到这一点?

Will not work. Any ideas how to do this ?

推荐答案

你可以指定匿名函数的类型，而不是像这样的参数类型:

You could specify type of anonymous function, instead of types of parameters like this:

val fun: (Boolean, => Int) => Int = (x, y) => if(x) y else 0

scala> fun(false, {println("!"); 2})
res1: Int = 0

scala> fun(true, {println("!"); 2})
!
res2: Int = 2

<代码>=>Int 不是正确的类型名称，它是方法声明或匿名函数类型的参数块中按名称参数的特殊语法.

=> Int is not a correct type name, it's a special syntax for by-name parameters in parameters block of method declaration or anonymous function type.

参见 SLS 4.6 函数声明和定义强>

ParamType ::= Type
            | ‘=>’ Type
            | Type ‘*’

如果您不想将匿名函数分配给变量，您可以使用如下类型推断:

In case you don't want to assign anonymous function to variable you could either use type inference like this:

xxx[Int, Boolean]{ (x, y) => y }

或者这样指定它的类型:

Or specify its type this way:

xxx({ (x, y) => y }: ((Int, => Boolean) => Boolean))

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