Getting a Type from TypeInfo / Getting Variant Type

[Simen Kjærås]

On Thursday, 11 January 2018 at 08:59:01 UTC, Chirs Forest wrote:
> I'm using std.variant.Variant to hold a value of unknown type 
> (not a string, could be a numeric type or a container holding 
> multiple numeric types). I'm trying to retrieve this value with 
> .get!T but I need the type to do that... .type gives me 
> TypeInfo, but that's not a type so I'm not sure how to get the 
> type without checking the TypeInfo against all possible types 
> (which would be a massive pain).

Since types are compile-time features, you can't just turn a 
TypeInfo into a type, sadly. However, provided you use Algebraic 
and not Variant, you have a limited type universe, and can 
iterate over the possibilities:

import std.variant;
import std.stdio;

template dispatch(alias Fn)
{
     auto dispatch(A)(A arg)
     if (A.AllowedTypes.length > 0)
     {
         static foreach (T; A.AllowedTypes)
         {
             if (typeid(T) == arg.type) {
                 return Fn(arg.get!T);
             }
         }
     }
}

void test(int n) {
     writeln("It's an int!");
}

void test(string s) {
     writeln("It's a string!");
}

unittest {
     Algebraic!(int, string) a = 4;
     a.dispatch!test(); // It's an int!
     a = "Test";
     a.dispatch!test(); // It's a string!

     Algebraic!(float, int[]) b = 2.3f;
     a.dispatch!test(); // Will not compile, since test has no 
overloads for float or int[].

     Variant v = 3;
     v.dispatch!test; // Will not compile, since we don't know 
which types it can take
}

--
   Simen