请为什么我的代码不起作用，我需要安装任何库

发布于 2025-01-29 00:22:28 字数 324 浏览 3 评论 0原文

最近，我开始使用功能编程和管道的所有解释，并使用我看到的减少构成非常粗略的。


const x = 4

const add2 = x + 2

const multiplyBy5 = x * 5

const subtract1 = x - 1



pipe = (...functions) =>
(x) => functions.reduce((v, function) => function(v), x)

const result = pipe(add2, multiplyBy5, subtract1)(4)
console.log(result)

原文

Recently started functional programming and all explanations of the pipe and compose using reduce which I have seen are very sketchy.


const x = 4

const add2 = x + 2

const multiplyBy5 = x * 5

const subtract1 = x - 1



pipe = (...functions) =>
(x) => functions.reduce((v, function) => function(v), x)

const result = pipe(add2, multiplyBy5, subtract1)(4)
console.log(result)

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场罚期间 2025-02-05 00:22:28

有2个错误。

第一个是x，add2，pultiplyby5 and 减去1不是功能，而是仅仅定义。
另一个是，您将变量（使用参数）命名为一个“保留”单词（例如“函数”）的名称确实打破了语法解析器。

const x = (x) => x
const add2 = (x) => x+2
const multiplyBy5 = (x) => x*5
const subtract1 = (x) => x-1

const pipe = (...functions) => (x) => functions.reduce((v,fn)=>fn(v),x)
const result = pipe(
  add2,
  multiplyBy5,
  subtract1,
)(4);
console.log(result)

There were 2 errors.

The first one was that the x, add2, multiplyBy5 and subtract1 were not functions, but mere definitions.
The other was that you naming a variable (using the arguments) to a name that is a "reserved" word such as "function" did break the syntax parser.

const x = (x) => x
const add2 = (x) => x+2
const multiplyBy5 = (x) => x*5
const subtract1 = (x) => x-1

const pipe = (...functions) => (x) => functions.reduce((v,fn)=>fn(v),x)
const result = pipe(
  add2,
  multiplyBy5,
  subtract1,
)(4);
console.log(result)

回复收藏 0 原文

初相遇 2025-02-05 00:22:28

我认为应该这样做：

const add2 = (x) => x+2
    
const multiplyBy5 = (x) => x*5
    
const subtract1 = (x) => x-1
    
const pipe=(...functions) => (x) => functions.reduce((v, functionA) => functionA(v), x)
    
const result=pipe(add2, multiplyBy5, subtract1)(4)
    
console.log(result)

I think it should be done like this:

const add2 = (x) => x+2
    
const multiplyBy5 = (x) => x*5
    
const subtract1 = (x) => x-1
    
const pipe=(...functions) => (x) => functions.reduce((v, functionA) => functionA(v), x)
    
const result=pipe(add2, multiplyBy5, subtract1)(4)
    
console.log(result)

回复收藏 0 原文

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