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
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