I have a parameterized class. I would like to get the name of the class represented by the class name. For instance, what I want to do is this:
public T foo(){
System.out.println(T.class.getName());
}
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stackoverflow.com/questions/3403909/… possible duplicatenarek.gevorgyan– narek.gevorgyan2012-01-23 14:10:50 +00:00Commented Jan 23, 2012 at 14:10
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Yes, duplicate, I agree. How do I fix this?Thom– Thom2012-01-23 14:12:45 +00:00Commented Jan 23, 2012 at 14:12
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3 Answers
You can't do it this way, since T isn't known at compile time. You could achieve something similar like so:
public void foo(T t) {
System.out.println(t.getClass().getName());
}
Note that this takes an instance of T and would print out the name of its dynamic type.
Whether or not this is a good enough substitute depends on your use case.
Java generics don't work that way. If you have any bounds on T, you can access the bounds by querying the type variable definition. E.g.:
public class Foo<T extends Bar>{}
will let you get at Bar, but not at the subtype of Bar you are actually using. It doesn't work, sorry.
Read the Java Generics FAQ for more info.
BTW: One common solution to this problem is to pass the subtype of T into your class, e.g.
public T foo(Class<? extends T> tType){
System.out.println(tType.getName());
}
I know it's cumbersome, but it's all Java generics allow.
answered Jan 23, 2012 at 14:10
Sean Patrick Floyd
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2 Comments
ziesemer
Actually, you can get what
T is. See my answer for how. (Not sure why it was downvoted!)Sean Patrick Floyd
@ziesemer you can only get at bounds that are actually there. If the above T was declared
T extends Something, you could get at the something part. But if your type variables have no bounds there's nothing to get at.public T foo(T t){
System.out.println(t.getClass().getName());
}
