Scala:返回对函数的引用 [英] Scala: return reference to a function

问题描述

我想要一个 Scala 函数来返回对另一个函数的引用，这可能吗?

I'd like to have a Scala functions that returns the reference to another function, is that possible?

推荐答案

您可以返回一个函数类型(这是由 A => B 定义的).在这种情况下，从整数到整数:

You can return a function type (this is defined by A => B). In this case Int to Int:

scala> def f(x:Int): Int => Int =  { n:Int => x + n }
f: (x: Int)(Int) => Int

当您调用该函数时，您将获得一个新函数.

When you call the function you get a new function.

scala> f(2)
res1: (Int) => Int = <function1>

可以作为普通函数调用的:

Which can be called as a normal function:

scala> res1(3)
res2: Int = 5

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