[] is an empty array ![] is treated as false combining a boolean with the empty array returns "false" as a string (so true + [] = "true", false + [] = "false") ! + [] is treated as true ! + [] + ! + [] is treated as 2 since true + true = 1 + 1 = 2 so you have "false"[2], which is l for the o it's [] is an empty array [] + {} returns "[object Object]" as a string ({} + [] returns 0) ![] is false !![] is true +!![] casts it to an integer so that part is "[object Object]"[1], which returns "o"

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[–] nulldev@programming.dev 136 points 2 years ago (3 children)

The explanation is about as understandable as the JS code, however thanks OP, TIL

[–] Deebster@lemmyrs.org 37 points 2 years ago* (last edited 2 years ago)

The line breaks haven't worked, here's it formatted correctly:

Explanation: it's mostly due to how js does type conversion.

For the Ls, it's:

[] is an empty array
![] is treated as false
combining a boolean with the empty array returns "false" as a string (so true + [] = "true", false + [] = "false")
! + [] is treated as true
! + [] + ! + [] is treated as 2 since true + true = 1 + 1 = 2
so you have "false"[2], which is l

for the o it's:

[] is an empty array
[] + {} returns "[object Object]" as a string ({} + [] returns 0)
![] is false
!![] is true
+!![] casts it to an integer
so that part is "[object Object]"[1], which returns "o"-

[–] Sonotsugipaa@lemmy.dbzer0.com 21 points 2 years ago

Iirc this is called "JSFuck", and it has a number of compilers

[–] elvith@feddit.de 6 points 2 years ago (1 children)

There's a nice presentation that explains this behaviour quite understandable: https://youtu.be/sRWE5tnaxlI

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