_占位符的scala类型推断 [英] scala type inference with _ place holder

问题描述

List("This","is","Scala").foreach(a => print(a+" "))

编译正常，但是

List("This","is","Scala").foreach(print(_+" "))

无法抱怨缺少参数类型.我不知道为什么它会失败.

fails complaining of missing parameter type. I couldn't figure out why it fails.

我的意思是打印而不是println-并不是说它在逻辑上有所不同.

I meant print not println - not that it makes logical difference.

推荐答案

问题是这个

List("This","is","Scala").foreach(print(_+" "))

不等同于

List("This","is","Scala").foreach(a => print(a+" "))

但要

List("This","is","Scala").foreach(print(a => a+" "))

现在，让我们来看一下foreach的类型签名:

Now, let's see the type signature of foreach:

def foreach [B] (f: (A) ⇒ B) : Unit

其中，A是List本身的类型参数.由于我们有一个List[String]，因此编译器知道必须将Function[String, B]传递给foreach.

where A is the type parameter of the List itself. Since we have a List[String], the compiler knows one has to pass to foreach a Function[String, B].

在a => print(a+" ")中，a的类型是已知的，然后:String.

In a => print(a+" ") the type of a is already known then: String.

在print(a => a+" ")中存在问题，因为print不是Function.但是，编译器尚未考虑-仍在尝试编译a => a+" ".因此，让我们来看一下Predef.print的类型:

In print(a => a+" ") there is a problem, as print is not a Function. However, the compiler hasn't considered that yet -- it's still trying to compile a => a+" ". So let's look at the type of Predef.print:

def print (x: Any) : Unit

因此，a => a+" "必须是Any类型，这当然意味着它可以是任何东西.它并没有帮助编译器断言a的类型是什么.其实并不重要，因为您不想首先打印Function.

So a => a+" " must be of type Any, which, of course, means it can be anything. It doesn't help the compiler in asserting what the type of a is. Which doesn't really matter, because you didn't want to print a Function in first place.

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