'Get trait in subclass
I'm sorry if this is poorly worded or if this has been asked before but I couldn't seem to find anything related to this and I'm quite tired.
Alright, so what I'm trying to do is get the value of of my trait in a subclass for situations where I need to reference an instance of a subclass but I don't have the information about what trait it will be using. This is easier for me to explain in code so here's what I'm trying to do.
public class TraitUser<T>
{
public void DoThingWithT(T thing)
{
thing.ToString();
}
}
public class TraitInspector
{
public void DoThing()
{
// This is where I run into my issue,
// I need to be able to get the trait that
// an instance of the TraitUser class is using to continue.
TraitUser<> tUser = GetRandomTraitUser()/*Imagine this returns an instance of TraitUser with a random trait, this is where my issue comes in.*/;
}
}
Solution 1:[1]
If I understand youright, you need get information about generic type T in TraitUser instance in TrairInspector.
public interface IGetTraitInfo
{
Type GetTraitObjectType();
object GetTraitObject();
}
public class TraitUser<T> : IGetTraitInfo
{
private T _thing;
public void DoThingWithT(T thing)
{
_thing = thing;
}
public Type GetTraintObjectType()
{
return typeof(T);
}
public Type GetTraitObject()
{
return _thing;
}
}
public class TrairInspector
{
public void InspectTraitUser(IGetTraitInfo traitUser)
{
Type traitType = traitUser.GetTraintObjectType();
object data = traitUser.GetTraitObject();
}
}
Solution 2:[2]
I didn't understand completely but this might help you.
public interface ITrait
{
string DoSomething();
}
public class Trait<T> where T : ITrait, new()
{
public string DoSomething()
{
ITrait trait = new T();
return trait.DoSomething();
}
}
public class TraitUser : ITrait
{
public string DoThing()
{
return "return something";
}
}
public class TrairInspector
{
public void DoThing()
{
Trait<TraitUser> traitUser = new Trait<TraitUser>();
traitUser.DoSomething();
}
}
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | opewix |
| Solution 2 | Nagaraja S |
