5 ms·

Try it letter by letter. For instance, to get the string "a": (![]+[])[+[[+!+[]]]] Take the first part, `(![]+[])`. `![]` evaluates to `false`. Then `+[]

by mjhoy 13y ago

Try it letter by letter. For instance, to get the string "a": (![]+[])[+[[+!+[]]]] Take the first part, `(![]+[])`. `![]` evaluates to `false`. Then `+[]` coerces false into a string, so the expression is `"false"`. The rest of the expression (more complicated) evaluates to `[[1]]`, which will grab the `"a"` from `"false"`. Now why there is the extra surrounding brackets, I'm not sure, because `[1]` would have worked as well.

bargl 13y ago
It looks like the generator is suboptimal, but still really cool. (![]+[])[+!+[]] evaluates to "a" as well. It looks like it turns it into an array twice. It goes +!+[] === 1 then [+!+[]] === [1] then +[[+!+[]]] === 1 then [+[[+!+[]]]] === [1]
- mjhoy 13y ago
  Ah! You're right. Confusing. Also confusing that "false"[[1]] works.
- ygra 13y ago
  Maybe one could run the output through Closure Compiler to get a bit better code ;-) (Ok, tried it, it evaluates plenty of numbers on its own already which increases the character set. Not nice).