如何转换 X =>Option[R] 到 PartialFunction[X,R] [英] How to convert X => Option[R] to PartialFunction[X,R]

问题描述

只要我们有一个 PartialFunction[X,R]，很容易将它转换成一个返回 Option[R] 的函数，例如

def pfToOptf[X, R](f: PartialFunction[X,R])(x: X) =if (f.isDefinedAt(x)) Some(f(x))其他 无

但是，如果任务相反怎么办:假设我有一个函数 f 获取 X 作为参数并返回 Option[R]因此.我想用它制作一个 PartialFunction[X,R].最好的方法是什么?

我想出的东西在我看来很丑陋:

def optfToPf[X,R](f: X => Option[R]) : PartialFunction[X,R] = {对象提取器{def unapply(x: X): Option[R] = f(x)}{案例提取器(r)=>r }}

我错过了更好的方法吗?

解决方案

Starting Scala 2.9, Function.unlift 正是这样做的:

def unlift[T, R](f: (T) => Option[R]): PartialFunction[T, R]

<块引用>

将函数 T => Option[R] 转换为 PartialFunction[T, R].

As long as we have a PartialFunction[X,R] it's very easy to convert it to a function returning Option[R], e.g.

def pfToOptf[X, R](f: PartialFunction[X,R])(x: X) =
    if (f.isDefinedAt(x)) Some(f(x))
    else None

However, what if the task is opposite: suppose I have a function f getting X as an argument and returning Option[R] as a result. And I want to make a PartialFunction[X,R] out of it. What is the best way?

What I've come up with looks pretty ugly to my taste:

def optfToPf[X,R](f: X => Option[R]) : PartialFunction[X,R] = {
    object extractor {
        def unapply(x: X): Option[R] = f(x)
    }

    { case extractor(r) => r }
}

Is there some better way I missed?

解决方案

Starting Scala 2.9, Function.unlift does precisely this:

def unlift[T, R](f: (T) => Option[R]): PartialFunction[T, R]

Turns a function T => Option[R] into a PartialFunction[T, R].

这篇关于如何转换 X =>Option[R] 到 PartialFunction[X,R]的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！