Inout unclearness

[Steven Schveighoffer via Digitalmars-d]

On 7/20/15 8:07 PM, Max Klimov wrote:
> On Monday, 13 July 2015 at 13:50:02 UTC, Steven Schveighoffer wrote:
>> With inout, it's even more tricky, because the compiler has to ensure
>> it's not inout when it leaves the function. And it's not possible in
>> some cases to do this.
>
>> You can simply use a template function without hand-written cases.
>> This is what inout was meant to replace.
>
> Sure, however, it is still not the same as using inout.
> Firstly, your function should be instantiated somewhere for mutable T,
> const T and immutable T. Otherwise, compilation errors will be postponed
> until the code usage.

Sure, you can do this via unit tests.

> Secondly, template function can not be virtual.

Unfortunately, this is true. I don't know how this could be fixed, and 
inout certainly does help in this regard.

> I'm wondering why it is needed to have special casting rules and other
> restrictions for inout if people should treat inout as wildcard for
> mutable, immutable and const.

Again, it has nothing to do with code generation, it has to do with type 
conversion. The compiler doesn't know how to convert 
Rebindable!(inout(T)) to something that can be returned from an inout 
function. We haven't figured out how to tell it what to do there, so it 
has to give up.

> As far as I can imagine, a compiler should
> generate one single object code for inout function. Is it possible just
> to check validity of an inout(T) function for T, const(T) and
> immutable(T) and then, if success, generate the code? (keeping in mind
> that you can execute another inout functions inside this one). Am I
> missing something?

It has to not only successfully generate code, the code itself has to be 
exactly the same for all 3 versions (and actually, if you have multiple 
inout parameters, it potentially has to run 3^parameter tests). It could 
potentially do this, but inout works just as effectively with the 
current rules. It still does not provide a way to convert a struct with 
an inout member to a struct without one.

-Steve