alert(
(![]+[])[[]-[]]+
(([]+[])+([][[]]))[[]-[]]+
(([]+[])+([][[]]))[!![]-[]]
);
Heres' the fiddle: http://jsfiddle.net/leeny/6VugZ/
How exactly is this cryptic piece of code working?
Asked
Active
Viewed 1,334 times
7
PhD
- 11,202
- 14
- 64
- 112
-
1Because strings can be indexed. – Dave Newton Mar 08 '13 at 20:43
-
Break it down in the console. It's awesome. – Austin Mullins Mar 08 '13 at 20:43
-
2Ugh; why is this getting upvotes? All you need to do, if you can't figure it out by looking at it, is try one small piece at a time. It's a little depressing. – Dave Newton Mar 08 '13 at 20:46
-
2The same way these work: http://stackoverflow.com/q/4170978/218196, http://stackoverflow.com/q/7202157/218196. – Felix Kling Mar 08 '13 at 20:50
1 Answers
7
vvvvvvv [0]
(![]+[])[[]-[]] = "false"[0]
^^^^^^^^ "false"
vvvvvvv again [0]
(([]+[])+([][[]]))[[]-[]] = "undefined"[0]
^^^^^^^^^^^^^^^^^^ "undefined"
vvvvvvvvv this time [1]
(([]+[])+([][[]]))[!![]-[]] = "undefined"[1]
^^^^^^^^^^^^^^^^^^ again "undefined"
Thus you get "f"+"u"+"n" === "fun".
Further explanation
"false"
![] is false. +[] simply acts as a transformation into a string. Thus we gain the string "false".
"undefined"
One of the operands needs to be a string. This is being done by []+[]. The actual undefined is created in the right hand side: [][[]] === [][0], the first entry of an empty array is undefined.
-
1As a side note, `(([]+[])+([][[]]))` seems like overkill to me, `([]+[][[]])` appears to suffice. – Neil Mar 08 '13 at 21:20
-
@Neil: There are other overkills too: `[[]-[]] === [+[]]`. I believe `(![]+[])[+[]]+([]+[][[]])[+[]]+([]+[][[]])[+!![]]` is the shortest sequence which creates the same result. – Zeta Mar 08 '13 at 21:24