在javascript中的高阶函数中使用原型函数

问题描述

我正在尝试使用reduce合并数组数组，我发现可以使用Array.prototype.concat函数，如下所示：

I'm trying to concat an array of arrays using reduce and I figured that I could use the Array.prototype.concat function like this:

烧入它的函数：

You can use bind to create a copy of a function with a particular this burned into it:
arr.reduce(Array.prototype.concat.bind(Array.prototype), [])
但是，现在已经清除，还有其他一些问题使您无法执行此操作。
However, now that that's cleared up, there are several other issues that stop you from doing this.
其中一个，实际上是减少获取四个参数，包括当前索引和整个数组。通过让您的（a，b）=>  lambda只将这四个参数中的两个传递给 concat 来忽略它们。很好，但是当您直接提供一个函数作为 reduce 的参数时，它将使用所有四个参数，因此您将得到调用<$ c $的结果c> Array.prototype.concat（a，b，currentIndex，arr）。
For one, reduce actually gets four arguments, including the current index and the whole array. You ignore these by having your (a,b)=> lambda only pass two of those four arguments into concat. That's fine, but when you supply a function directly as an argument to reduce, it will use all four arguments, so you'll get the result of the call Array.prototype.concat(a, b, currentIndex, arr).
此外，您正在做的不是明智地使用 Array.prototype 。  concat 函数连接其参数，并将其附加到 this 值的副本中。由于 Array.prototype 本身只是一个空数组（尽管具有许多其他数组用作继承属性的自有属性），所以这实际上与 []。concat（a，b）或（也许更可读） a.concat（b）。
Furthermore, what you're doing isn't a sensible use of Array.prototype. The concat function concatenates its arguments and appends them to a copy of the this value. Since Array.prototype is itself just an empty array (albeit with many own-properties that other arrays use as inherited properties), this is effectively the same as [].concat(a,b) or (perhaps even more readably) a.concat(b).
                        
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