如果返回的函数在别处声明，则柯里化不起作用

Question

以下代码运行良好：

// Returns a function that inherits parent's scope
const curry = argOne => argTwo => console.log(argOne, argTwo);

const initiateFunction = curry(1);
initiateFunction(2) // Outputs: "1 2"

我的意图是在代码的其他地方声明一个返回的函数以使其可重用。

尽管如此，如果我写：

const myLogger = argTwo => console.log(argOne, argTwo);

// Returns reference to a function that doesn't inherit parent's scope
const curry = argOne => myLogger;

const initiateFunction = curry(1); // Seems to be ok

initiateFunction(2); // ReferenceError: argOne is not defined

有什么问题以及如何解决？

Answer 1

我很困惑，因为那时柯里化对我来说是一个非常新的概念。结合箭头函数，看起来很棒，但也更难掌握。

如果我们使用经典的函数定义重写这个例子，就很容易理解为什么第二种方式不起作用。

const curry = argOne => argTwo => console.log(argOne, argTwo);
curry(1)(2) // -> 1 2

// Is equivalent to
function curry (argOne) {
  return function(argTwo) {
    console.log(argOne, argTwo)
  }
}

而在第二个示例中，myLogger 无法访问 argOne，如评论中正确所述：

const myLogger = argTwo => console.log(argOne, argTwo);
const curry = argOne => myLogger;

curry(1)(2) // -> Error

// Is equivalent to
function myLogger(argTwo) {
  console.log(argOne, argTwo)
}

function curry(argOne) {
  return myLogger // which has no idea what is argOne since declared outside of curry scope
}

如果返回的函数在别处声明，则柯里化不起作用

Currying doesn't work if returned function declared elsewhere

javascript

closures

currying