Scala 中的柯里化示例 [英] Currying Example in Scala

问题描述

以下是柯里化的好例子吗?

Is the following a good example of currying?

def sum(a: Int, b: Int) : (Int => Int) = {
    def go(a: Int) : Int = {
        a + b;
    }
    go
}

我对下面的结果有点理解，但是我怎么能用咖喱的方式写(或者我应该怎么写)sum()?

I half understand the below results, but how could I write (or maybe how I should've written) sum() in a curried way?

scala> sum(3,4) res0: Int => Int = <function1>
scala> sum(3,4).apply(2) res1: Int = 6
scala> sum(3,4).apply(3) res2: Int = 7

推荐答案

在 Scala 中引入了柯里化机制来支持类型推断.例如标准库中的 foldLeft 函数:

Currying mechanism was introduced in Scala to support type inference. For example foldLeft function in the standard lib:

def foldLeft[B](z: B)(op: (B, A) => B): B

在不柯里化的情况下，您必须明确提供类型:

Without currying you must provide types explicitly:

def foldLeft[B](z: B, op: (B, A) => B): B
List("").foldLeft(0, (b: Int, a: String) => a + b.length)
List("").foldLeft[Int](0, _ + _.length)

柯里化函数的三种写法:

There are three ways to write a curried function:

1) 写成柯里化形式:

1) Write it in currying form:

def sum(a: Int)(b: Int) = a + b

这只是语法糖:

def sum(a: Int): Int => Int = b => a + b

2) 在函数对象 (sum _).curried 上调用 curried 并检查类型:

2) Call curried on the function object (sum _).curried and check the types:

sum: (a: Int, b: Int)Int
res10: Int => (Int => Int) = <function1>

在您的示例中，您可以使用 Scala 类型推断来减少代码量并更改您的代码:

In your example, you can use Scala type inference to reduce the amount of code and change your code:

def sum(a: Int, b: Int) : (Int => Int) = {
    def go(a: Int) : Int = {
        a + b;
    }
    go
}

进入:

def sum(a: Int, b: Int) : (Int => Int) = c => a + b + c

语义上这些是相同的，因为您明确提供了返回类型，因此 Scala 知道您将返回一个带有 Int 参数的函数并返回一个 Int

semantically these are the same, because you explicitly provided the return type, so Scala knows that you will return a function wich takes an Int argument and return an Int

关于curring的更完整的答案是由retronym

Also a more complete answer about curring was given by retronym

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