How to avoid inout type constructor with Optional type wrapper undoing string type

[Jacob Carlborg]

On 2018-07-23 20:39, aliak wrote:
> Hi,
> 
> I'm playing around with an Optional wrapper type. It stores a type T and 
> a bool that defines whether a value is defined or not:
> 
> struct Optional(T) {
>    T value;
>    bool defined = false;
>    this(U : T)(auto ref inout(U) value) inout {
>      this.value = value;
>      this.defined = true;
>    }
> }
> 
> To facilitate it's use I have two type constructors:
> 
> inout(Optional!T) some(T)(auto ref inout(T) value) {
>      return inout(Optional!T)(value);
> }
> 
> Optional!T no(T)() {
>      return Optional!T();
> }
> 
> The above produces a problem when working with strings. Basically the 
> type information gets slightly altered so you can't do this:
> 
> auto a = [no!string, some("hello")];
> 
> You get a type mismatch:
> 
> * no!string = Optional!string
> * some("hello") = immutable(Optional!(char[]))
> 
> I've created a short code gist, so basically I'm wondering how to get it 
> to compile without changing what's in main()
> 
> https://run.dlang.io/is/BreNdZ
> 
> I guess I can specialize on string type T, but this is a more general 
> problem that can be shown with:
> 
> struct S {}
> alias Thing = immutable S;
> Thing thing = S();
> 
> auto x = some(thing);
> auto y = no!Thing;
> auto arr = [x, y]; // no can do buddy

This [1] compiles the first example but not the second.

[1] https://run.dlang.io/is/SJ02kP

-- 
/Jacob Carlborg