Scala类型语法 [英] Scala Type Syntax

问题描述

我观察到，如果我想创建一个可以接受任何类型列表并返回布尔值的泛型函数，那么我可以使用以下语法来进行函数声明：

  def someFunction [A]（l：List [A]）：Boolean 
  

然而，我可以用这个语法实现一个等价的函数声明：

  def someFunction （l：List [_]）：布尔
  

后面的语法对我来说很有意义;下划线表示任何类型的List的通配符。然而前者令人困惑;这两种语法之间的语义差异是什么？注意：我注意到在第一个语法示例中，我可以使用[B]或[c]甚至[％]代替[A]。

解决方案

A 是一个类型参数。就像一个值参数，比如你的 l 传递参数，它就是某个类型的名称或占位符，它可能在不同的时间是不同的。在这个例子中， A 没有被使用，所以，是的，使用<$ <

c $ c> _ 更有意义并且更清晰，但是如果要从列表中返回一个元素，那么方法返回类型将是 A （或任何你想给这个参数的名字）。使用 _ 作为返回类型没有任何意义。

I've observed that, if I want to make a generic function that can accept a list of any type and return a boolean, I can use the following syntax for a function declaration:

def someFunction[A](l:List[A]):Boolean

However, I can achieve an equivalent function declaration with this syntax as well:

def someFunction(l:List[_]):Boolean

The latter syntax makes sense to me; the underscore indicates a wildcard for a List of any type. However the former is confusing; what's the semantic difference between the two types of syntax, if there is one at all? Note: I noticed I could use [B] or [c] or even [%] in place of the "[A]" in the first syntax example.

解决方案

The A is a "type parameter". Just like a value parameter, such as your l passed parameter, it is the "name", or place holder, for a some type which might be different at different times (i.e. with different invocations of the method).

In your example the A is unused so, yeah, using _ makes more sense and is clearer, but if you were to return an element from the list then the method return type would be A (or whatever name you want to give that parameter). Using _ as a return type wouldn't make any sense.

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