为什么JS中的空数组加false会返回字符串? [英] Why does an empty array plus false in JS return a string?

问题描述

为什么一个空数组加false会返回字符串"false"?

Why does an empty array plus false return the string "false"?

> [] + false

> "false"

一个空数组是假的，对吧?

An empty array is false, right?

然后，假+假=假?不?

Then false + false = false? No?

推荐答案

简短答案:因为

Short answer: because the spec says so.

更长的答案:当 + 运算符的左操作数不是字符串或数字时，该操作数将根据其首选"原始类型(定义为规格).数组的首选"类型是字符串，而 [].toString()是空字符串.

Longer answer: When the left operand of the + operator is not a string or a number, the operand will be converted to either one of those depending on their "preferred" primitive type (defined in the spec). The array's "preferred" type is a string and [].toString() is an empty string.

然后，仍然按照规范，因为左操作数是一个字符串(在上一次转换之后)，所以右操作数将被转换为字符串，并且 + 操作的结果为两个字符串的串联.

Then, still according to the spec, because the left operand is a string (following the previous conversion), the right operand will be converted to a string and the result of the + operation will be the concatenation of both strings.

换句话说， [] + false 等效于 [].toString()+ false.toString()(或" +"false")，并生成字符串" false".

In other words, [] + false is equivalent to [].toString() + false.toString() (or "" + "false") and results in the string "false".

由此产生的其他有趣结果:

Other interesting results as a consequence of this:

[1,2,3] + false // "1,2,3false"
[1,[2]] + false // "1,2false"
[] + {} // "[object Object]"

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