4 ms·
> unless you're appending |0 to everything like a compiler would, i.e. ((a|0) / (b|0))|0. It effectively does do that though, since earlier in your code `a` an
by Asraelite 4y ago
> unless you're appending |0 to everything like a compiler would, i.e. ((a|0) / (b|0))|0.
It effectively does do that though, since earlier in your code `a` and `b` would be defined with `let a = ... |0` so now `a` and `a|0` are equivalent in any subsequent expressions.
- chrisseaton 4y agoYes - local variables are just indirections through a name - in a compiler referring to a and referring to the original ... |0 expression are the same thing. https://twitter.com/ChrisGSeaton/status/1530579735012593665 https://twitter.com/ChrisGSeaton/status/1530579735012593665
- deleted 4y ago[deleted]