Typedef!int + Typedef!int => int? is this a typedef overlook, or it's a feature by design?

[RazvanN]

On Wednesday, 10 June 2020 at 22:47:49 UTC, mw wrote:
> Hi,
>
> I'm trying to follow this doc on typedef:
>
> https://dlang.org/library/std/typecons/typedef.html
>
> --------------------------------
> import std.typecons;
> import std.stdio;
>
> alias MyInt = Typedef!int;
>
> void f(MyInt mi) {}
>
> void main() {
>   MyInt a = 2;
>   MyInt b = 3;
>   f(a + b);
> }
> --------------------------------
>
> td.d(14,4): Error: function td.f(Typedef!(int, 0, null) mi) is 
> not callable using argument types (int)
> td.d(14,4):        cannot pass argument a.opBinary(b) of type 
> int to parameter Typedef!(int, 0, null) mi
>
> Naturally I've expected `a + b` will have the same type as 
> MyInt (not the underlying type int).
>
> Is this a typedef overlook, or it's a feature by design?
>
> If it's by design what's the reason behind?
>
> Thanks.

I think it is an overlook and therefore a bug. The spec says that 
the whole point of Typedef is to create a type that is different 
from the typedefed one (otherwise you would end up with a 
glorified alias). However the implementation simply mixins a 
generic implementation of opBinary that uses the underlying type 
[1]. The definition of opBinary has no idea that it should wrap 
the result in another type [2].

[1] 
https://github.com/dlang/phobos/blob/master/std/typecons.d#L7472
[2] 
https://github.com/dlang/phobos/blob/master/std/typecons.d#L6836