该“高阶函数"如何进行?事情在Javascript中有效 [英] How does this "higher-order functions" thing works in Javascript

问题描述

在Marijn Haverbeke的 Eloquent Java 一书中，有此示例在介绍高阶函数的概念时:

From the book Eloquent Javascript by Marijn Haverbeke, there is this example while introducing the concept of higher-order functions:

function greaterThan(n) {
  return function(m) { return m > n; };
}
var greaterThan10 = greaterThan(10);
console.log(greaterThan10(11));
// → true

我不太确定这是如何工作的……可能回答了我自己的问题，但这是我的看法:

I'm not quite sure how this works... probably answering my own question, but this is how I see it:

• 首先，在此行中调用greaterThan(n)，并将其值分配给greaterThan10变量:

• First, greaterThan(n) is called in this line, assigning its value to the greaterThan10 variable:

var greaterThan10 = greaterThan(10);

这使存储为greaterThan10的函数看起来像:

This makes the function stored as greaterThan10 looks like:

function greaterThan(10) {
  return function(m) { return m > 10; };
}

然后，当您调用greaterThan10(11)时，您正在调用上面的函数，该函数转换为:

Then, when you call greaterThan10(11) you are calling the function above, which translates to:

function greaterThan(10) {
  return function(11) { return 11 > 10; };
}

因此，返回True作为11 > 10的结果确实是正确的.

Hence returning True as the result as 11 > 10 is true indeed.

有人可以确认我是否正确吗?另外，如果有人可以提供有关此高阶函数如何在JavaScript中工作的更多详细信息和评论，将不胜感激.

Could someone confirm whether I'm correct or not? Also, if someone can provide further details and comments on how this higher-order functions work in JavaScript, that would be greatly appreciated.

推荐答案

从正确的角度来看，您是正确的，但对它的评估略有不同.

You're correct, from a level of understanding, but it's evaluated slightly differently.

var greaterThan10 = greaterThan(10);

此行不会使存储为greaterThan10的函数看起来像任何东西-它创建了一个新函数，将变量n传递给它，因此greaterThan10变成了一个看起来像<的函数/p>

This line doesn't make the function stored as greaterThan10 "look like" anything - it creates a new function, passing in the variable n to it, so that greaterThan10 becomes a function that looks like

var greaterThan10 = function(m) { return m > 10; };

当您调用它时，您是在直接调用此函数，而不再需要遍历原始函数.

When you call it, you are calling this function directly, not going through the original function at all anymore.

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