On Wed, Apr 10, 2013 at 3:37 PM, Chad Zawistowski <chadz....@gmail.com>wrote:
> In the case where both declarations are on the same line, I agree with the > results of Niko's straw poll. > > let mut foo = 1, bar = 2; > > This should make both foo and bar be mut, in my opinion. > > Chad > The question is whether mut applies to let or to foo: (let mut) foo, bar vs. let (mut foo, bar). Is there any other place left in the language besides let, @, and & where mut is allowed? The latter cases seem to support (let mut): if you have &mut foo, it's a pointer through which mutation is allowed, named foo, in other words (&mut) foo, while the other way, &(mut foo), doesn't make any sense: it's not foo that's mutable! Similarly for @. So yeah, +1 for existing interpretation. -- Your ship was destroyed in a monadic eruption.
_______________________________________________ Rust-dev mailing list Rust-dev@mozilla.org https://mail.mozilla.org/listinfo/rust-dev
