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
