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