无法了解Scala函数应用程序中的占位符_行为 [英] Unable to understand placeholder _ behavior in scala function application

问题描述

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scala> val fn = (x:Int) => x+1
fn: Int => Int = <function1>

scala> val fn1 = fn _
fn1: () => Int => Int = <function0>

scala> val fn2 = fn1 _
fn2: () => () => Int => Int = <function0>

我不明白为什么函数的占位符(没有建议的类型)应用程序会创建一个带有额外的void参数前缀的新的curried函数.

I don't understand why the placeholder(without a suggested type) application to a function is creating a new curried function with prefixed additional void argument.

我期待编译器错误或至少警告.

I was expecting a compiler error or at least a warning.

推荐答案

我想是由于

I guess it's so due to the uniform access principle: in REPL val is a field of object, not a local variable. And all not private[this] fields are getter methods.

所以您的代码是这样的:

So your code is something like this:

def fn() = (x:Int) => x+1
val fn1 = fn _ // () => fn()

它可以与局部变量一起正常工作

It works as expected with local variables:

scala> {
     |   val fn = (x:Int) => x+1
     |   val fn1 = fn _
     | }
<console>:10: error: _ must follow method; cannot follow Int => Int
                val fn1 = fn _
                          ^

尽管我可以解释为什么这样做，但我仍然认为这种行为是错误的.

Even though I can explain why it works this way, I still think this behavior can be considered a error.

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