That would require an unnecessary move. It needs to be built into the language.
Eric Reed <[email protected]> wrote:
>I think the idea was to have the syntax desugar into method calls just
>like
>other existing operators.
>There'd be a trait like:
>
>trait Box<T> {
> fn box(val: T) -> Self
>}
>
>and something like "box expr in place" would desugar into
>"place::box(expr)".
>
>One question this poses is why are we requiring the "place" to be
>specified
>all the time?
>Why not let type inference handle deciding the "place" most of the
>time?
>
>
>On Mon, Dec 2, 2013 at 10:46 AM, Erick Tryzelaar
><[email protected]>wrote:
>
>> Is there any way we can use a method and move semantics for this?
>This
>> feels pretty natural to me:
>>
>> let foo = gc_allocator.box(bar);
>>
>>
>> On Mon, Dec 2, 2013 at 10:23 AM, Patrick Walton
><[email protected]>wrote:
>>
>>> Anything with @ feels like it goes too close to pointer sigils for
>my
>>> taste.
>>>
>>> Patrick
>>>
>>> spir <[email protected]> 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
>>>> [email protected]
>>>> https://mail.mozilla.org/listinfo/rust-dev
>>>>
>>>>
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>>>
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>>
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