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送给产品经理一段代码 (!(~+[]) + {})[--[~+""][+[]] * [~+[]] + ~~!+[]] + ({} + [])[[~!+[]] * ~+[]] #10

@xiaoyu2er

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@xiaoyu2er

送给产品经理一段代码 让他放到 console 去

(!(~+[]) + {})[--[~+""][+[]] * [~+[]] + ~~!+[]] + ({} + [])[[~!+[]] * ~+[]]

其实这段代码是我的同事发给我的, 我定睛一看一定有坑, 于是准备破解一番

其实这里面涉及到的知识点无非三个

  • 数据类型的转换
  • 位运算(按位非)
  • 运算符的优先级

分析之前我推荐大家看几篇文章

首先我们把代码进行拆分

(!(~+[]) + {})[--[~+""][+[]] * [~+[]] + ~~!+[]]
+
({} + [])[[~!+[]] * ~+[]]

简单来看就是
(A)[B] + (C)[D]

首先来看 A !(~+[]) + {}

+[] -(数据类型转换)-> 0
~+[] --> ~0 -(位运算)-> -1
!(~+[]) --> !(-1) -(类型转换)-> false
(!(~+[]) + {}) --> false + {} -(类型转换)-> 'false[object Object]'

再来看 B --[~+""][+[]]*[~+[]] + ~~!+[]


[~+""] --> [~0] --> [-1]
+[] --> 0
--[~+""][+[]] --> --[-1][0] --> --(-1) --> -2

[~+[]] --> [~0] --> [-1]

~~!+[] --> ~~!0 --> ~~true --> ~-2 -> 1

B --> -2 * [-1] + 1 --> 2 + 1 --> 3

那么 (A)[B] --> 'false[object Object]'[3] --> 's'

再来看 C ({} + [])

这里的 {} 其实是个代码块

所以等价于 +[][object Object]

再来看 D [~!+[]] * ~+[]

[~!+[]] --> [~!0] --> [!1] --> -2
~+[] --> ~0 --> -1
D -> -2 * -1 --> 2

所以 (C)[D] --> '[object Object]'[2] --> 'b'

那么

(!(~+[]) + {})[--[~+""][+[]] * [~+[]] + ~~!+[]] + ({} + [])[[~!+[]] * ~+[]]
--> (A)[B] + (C)[D) 
--> 's' + 'b' 
--> 'sb'

谢谢观看 😄

现在可以发给产品经理了

让他打开 chrome, (windows: F12, mac: command+shift+j)

粘贴 (!(~+[])+{})[--[~+""][+[]]*[~+[]] + ~~!+[]]+({}+[])[[~!+[]]*~+[]]

这就是我们前端一直想对你说的话

参考文章

文中对产品经理的冒犯仅为了行文效果 😅 前端和产品是相亲相爱的一家人 😆

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