[[][[]]+[]][+[]][++[+[]][+[]]] is "n" in javascript

[[][[]]+[]][+[]][++[+[]][+[]]]

This evaluates to "n" in javascript. Why?

Let's start with an empty array

[]

Now, let's access a member of it.

[][]

What member? Let's check for the empty array member

[][[]]

oh, that is undefined. But if we add an empty array to that, it is casted to the string "undefined"

[][[]]+[]

Let us wrap that in an array

[[][[]]+[]]

We can now try to access letters in that string. First, we must unwrap the string. That can be done by accessing the first element of that array.

[[][[]]+[]][0]

0 can be created by casting an empty array to a number:

[[][[]]+[]][+[]]

Now, "n" is the second letter in that string, so we would like to access that:

[[][[]]+[]][+[]][1]

But how can we write 1? Well, we increment 0, of course. Wrap 0 in an array, and increment the first member of it:

++[0][0]

Like before, this is equivalent to

++[+[]][+[]]

So our final code is then the glorious

[[][[]]+[]][+[]][++[+[]][+[]]][[][[]]+[]] [+[]][++[+[]] [+[]]]这在javascript中评估为“n”。 为什么？我们从一个空数组开始[]现在，让我们访问它的成员。[] []什么成员？ 我们来检查一下空的数组成员[] [[]]哦，这是未定义的。 但是，如果我们添加一个空数组，它被转换为字符串“undefined”[] [[]] + []让我们把它包装在一个数组中[[][[]]+[]]我们现在可以尝试访问该字符串中的字母。 首先，我们必须打开字符串。 这可以通过访问该数组的第一个元素来完成。[[][[]]+[]] [0]可以通过将一个空数组转换为数字来创建0：[[][[]]+[]] [+[]]现在，“n”是该字符串中的第二个字母，因此我们想访问：[[][[]]+[]] [+[]] [1]但是我们怎么能写1？ 那么当然我们增加0。 在数组中包装0，并增加其中的第一个成员：++[0] [0]像以前一样，这相当于++[+[]] [+[]]那么我们的最终代码就是光荣的[[][[]]+[]] [+[]][++[+[]] [+[]]]