Jonathan Blow

is this because you can then take advantage of all register bits being set to 1 (e.g. in simd code)?

Not only for simd, it's much, much more consistent behavior throughout. E.g. if `true` is signed, then `~true` is `false` in promoted integer arithmetic. With unsigned `~true` and `false` are different values.

cc @AbnerCoimbre because i don't know how Jai treats this

hm... so if bool was specified at the language level to be a signed int (and two's complement), then for any boolean b: ~b == !b ... which means you can get rid of either ~ or ! (and potentially use that one-token character for something else)

*er, "one-character token"

cc @Jonathan_Blow because now i'm even more curious about how Jai treats this

Bool is a separate type from integer with no implicit conversion. If you explicitly cast, true == 1. Whereas I can see some arguments for the all-1s case, I also can think of arguments for why 'true' should be 0, so, uhh, yeah.

You can't get rid of ! because !3 == false, but ~3 == ~3.