在Scala中指定Lambda返回类型 [英] Specifying the lambda return type in Scala

问题描述

注意:这是一个理论问题，我不是在试图解决任何问题，也不是为了实际目的而试图达到任何效果

使用(arguments)=>expression语法在Scala中创建lambda时，可以显式提供返回类型吗?

Lambda与方法没有什么不同，它们都被指定为表达式，但是据我所知，方法的返回类型很容易用def name(arguments): return type = expression语法定义.

考虑以下(说明性)示例:

def sequence(start: Int, next: Int=>Int): ()=>Int = {
    var x: Int = start

    //How can I denote that this function should return an integer?    
    () => {
        var result: Int = x
        x = next(x)
        result
    }
}

解决方案

您始终可以通过附加:和类型来声明表达式的类型.因此，例如:

((x: Int) => x.toString): (Int => String)

例如，如果您有一个复杂的大型表达式，并且不想依靠类型推断来使这些类型变得直截了当，那么这很有用.

{
  if (foo(y)) x => Some(bar(x))
  else        x => None
}: (Int => Option[Bar])
// Without type ascription, need (x: Int)

但是，如果将结果分配给具有指定类型的临时变量，则可能更加清楚:

val fn: Int => Option[Bar] = {
  if (foo(y)) x => Some(bar(x))
  else        _ => None
}

Note: this is a theoretical question, I am not trying to fix anything, nor am I trying to achieve any effect for a practical purpose

When creating a lambda in Scala using the (arguments)=>expression syntax, can the return type be explicitly provided?

Lambdas are no different than methods on that they both are specified as expressions, but as far as I understand it, the return type of methods is defined easily with the def name(arguments): return type = expression syntax.

Consider this (illustrative) example:

def sequence(start: Int, next: Int=>Int): ()=>Int = {
    var x: Int = start

    //How can I denote that this function should return an integer?    
    () => {
        var result: Int = x
        x = next(x)
        result
    }
}

解决方案

You can always declare the type of an expression by appending : and the type. So, for instance:

((x: Int) => x.toString): (Int => String)

This is useful if you, for instance, have a big complicated expression and you don't want to rely upon type inference to get the types straight.

{
  if (foo(y)) x => Some(bar(x))
  else        x => None
}: (Int => Option[Bar])
// Without type ascription, need (x: Int)

But it's probably even clearer if you assign the result to a temporary variable with a specified type:

val fn: Int => Option[Bar] = {
  if (foo(y)) x => Some(bar(x))
  else        _ => None
}

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