How to avoid inout type constructor with Optional type wrapper undoing string type
aliak
something at something.com
Mon Jul 23 18:39:59 UTC 2018
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
Cheers,
- Ali
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