DIP69 - Implement scope for escape proof references

[John Colvin via Digitalmars-d]

On Thursday, 11 December 2014 at 13:55:55 UTC, Marc Schütz wrote:
> On Thursday, 11 December 2014 at 12:48:05 UTC, Manu via 
> Digitalmars-d wrote:
>> On 8 December 2014 at 07:29, Walter Bright via Digitalmars-d
>> <digitalmars-d at puremagic.com> wrote:
>>> On 12/7/2014 6:12 AM, Dicebot wrote:
>>>>
>>>> But from existing cases it doesn't seem working good enough. 
>>>> For example,
>>>> not
>>>> being able to represent idiom of `scope ref int foo(scope 
>>>> ref int x) {
>>>> return x;
>>>> }` seems very limiting.
>>>
>>>
>>>  scope ref int foo(ref int x);
>>>
>>> will do it.
>>
>> Will it? It looks like foo can't be called with scope data?
>
> This is a point that most people don't seem to understand yet, 
> and which wasn't obvious for me either, at the beginning:
>
> * A `ref` parameter means that it cannot escape the function, 
> _except_ by return.
> * A `scope ref` parameter means that it cannot escape the 
> function  _ever_, not even by return.
> * A `scope ref` return means that it cannot leave the current 
> statement.
>
> Therefore, a `scope ref` return value can be passed on to the 
> next function as a `ref` argument. If that function again 
> returns a reference (even if not explicitly designated as 
> `scope`), the compiler will treat it as if it were `scope ref`.
>

OH! I had totally misunderstood that. Cheers for the explanation.