• Chirs Forest (7/7) Jan 11 2018 I'm using std.variant.Variant to hold a value of unknown type
• Jonathan M Davis (12/19) Jan 11 2018 If you have a variable of the type you put in, then you can use the type...
• Simen =?UTF-8?B?S2rDpnLDpXM=?= (40/47) Jan 11 2018 Since types are compile-time features, you can't just turn a

Chirs Forest <CF chrisforest.com> writes:

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).

Jonathan M Davis <newsgroup.d jmdavisprog.com> writes:

On Thursday, January 11, 2018 08:59:01 Chirs Forest via Digitalmars-d-learn 
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).

If you have a variable of the type you put in, then you can use the typeof
operator on it to get the type. e.g.

int i = 42;
static assert(is(typeof(i) == int));

It's even possible to declare variables of voldemort types that way - e.g.
if you want to save the result in a member variable.

But from what I can tell, you're either going to need to know the type you'r
dealing with, or you're going to need a variable of the type so that you can
then use typeof on it to get its type.

- Jonathan M Davis

Simen =?UTF-8?B?S2rDpnLDpXM=?= <simen.kjaras gmail.com> writes:

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