这可以重构使用泛型功能原理？ [英] Can this be refactored to use generic functional principles?

问题描述

比较函数升序接受两个参数 - a 和 b 。它必须返回一个比较两个整数的整数。

我有一个我想按名称排序的列表，所以我写了下面的函数。

是否有一个可用于组合这两个函数的函数式习惯用法，而不是让 byName 负责撰写得到的函数？

  const ascending =（a，b）=> a.localeCompare（b）中; 
 const byName =（i）=> i.get（姓名）; 
 const useTogether =（... fns）=> ...; //是否有像这样的惯用功能？ 
 
 // usage 
 items.sort（useTogether（byName（ascending）））; 
  

解决方案

我不确定它是否满足您的要求，重新寻找，但这里有一个可能的表达方式，你的 useTogether （有一些不同的签名）可以用于WORK。我不知道一个标准函数的效果。

const上升=（a，b）=> a.localeCompare（b）; const byName =（i）=> i ['name']; const useTogether =（selector，consumer）=> （... fnArgs）=> consumer（... fnArgs.map（selector））; var items = [{name：C}，{name：A}，{name：B}]; console.log（items.sort（ useTogether（byName，ascending）））

A comparator function ascending accepts two arguments - a and b. It must return an integer comparing the two.

I have a list that I want to sort by name, so I wrote the following functions.

Is there a functional idiom I can use to combine these two functions, rather than having byName take responsibility for composing the resulting function?

const ascending = (a, b) => a.localeCompare(b);
const byName = (i) => i.get('name');
const useTogether = (...fns) => ...; // is there an idiomatic function like this?

// usage
items.sort(useTogether(byName(ascending))); 

解决方案

I'm not sure if it satisfies what you're looking for, but here is one possible formulation of your useTogether (with a somewhat different signature) that does WORK. I'm not aware of a standard function with precisely that effect.

const ascending = (a, b) => a.localeCompare(b);
const byName = (i) => i['name'];
const useTogether = (selector, consumer) => (...fnArgs) => consumer(...fnArgs.map(selector));

var items = [{ name: "C" }, { name: "A" }, { name: "B" }];

console.log(
  items.sort(useTogether(byName, ascending))
)

这篇关于这可以重构使用泛型功能原理？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！