Inout unclearness

[Max Klimov via Digitalmars-d]

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.
Secondly, template function can not be virtual.

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. 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?

-Max