Explicitly saying ref or out when invoking a function

[Andrei Alexandrescu]

Lionello Lunesu wrote:
> 
> "Andrei Alexandrescu" <SeeWebsiteForEmail at erdani.org> wrote in message 
> news:h5ttfi$cf3$1 at digitalmars.com...
>> Lionello Lunesu wrote:
>>>
>>> "Jeremie Pelletier" <jeremiep at gmail.com> wrote in message 
>>> news:h5sl4c$1733$1 at digitalmars.com...
>>>> It's especially bad since if you modify the function prototype and 
>>>> change ref, you have all your calls to update too.
>>>
>>> That must be the best argument to introduce repeating ref and out!
>>>
>>> L.
>>
>> // untested
>> struct Ref(T)
>> {
>>     private T* _data;
>>     this(ref T data) { _data = &data; }
>>     ref T get() { assert(_data); return *_data; }
>>     alias get this;
>> }
>> Ref!(T) byRef(ref T data)
>> {
>>     return Ref!(T)(data);
>> }
>>
>> ...
>>
>> void fun(Ref!int rint) { ... }
>> ...
>> int x;
>> fun(byRef(x));
> 
> Andrei, you bring the compiler vs library discussion to a whole new 
> level :)
> 
> We can add "ref" to the list of keywords that can be freed up.

Good point, but actually ref can't be freed, because it would fall prey 
to infinite regression: if there was no ref keyword, how would you pass 
by reference the argument to the byRef function?

Andrei