But it's nothing like a pointer sigil. @ doesn't appear in the types at all. It's just the placement allocation operator.
On Mon, Dec 2, 2013 at 10:23 AM, Patrick Walton <pwal...@mozilla.com> wrote: > Anything with @ feels like it goes too close to pointer sigils for my > taste. > > Patrick > > spir <denis.s...@gmail.com> wrote: >> >> On 12/02/2013 11:57 AM, Kevin Ballard wrote: >> >>> With @ going away another possibility is to leave ~ as the normal >>> allocation operator and to use @ as the placement operator. So ~expr stays >>> the same and placement looks either like `@place expr` or `expr@place` >> >> >> I like that, with expr@place. Does this give: >> let foo = ~ bar; >> let placed_foo = bar @ place; >> ? >> >> Yet another solution, just for fun, using the fact that pointers are >> supposed to >> "point to": >> >> let foo = -> bar; >> let placed_foo = bar -> place; >> >> Denis >> ------------------------------ >> >> Rust-dev mailing list >> Rust-dev@mozilla.org >> https://mail.mozilla.org/listinfo/rust-dev >> >> > -- > Sent from my Android phone with K-9 Mail. Please excuse my brevity. > > _______________________________________________ > Rust-dev mailing list > Rust-dev@mozilla.org > https://mail.mozilla.org/listinfo/rust-dev > >
_______________________________________________ Rust-dev mailing list Rust-dev@mozilla.org https://mail.mozilla.org/listinfo/rust-dev
