digitalmars.D.learn - Getting a Type from TypeInfo / Getting Variant Type
- 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
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).
Jan 11 2018
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
Jan 11 2018
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
Jan 11 2018