Creating two generic parameters in an interface

Question

I am trying to create a generic interface for my classes that will satisfy them all. I have created base classes to link them all together but the interface I'm trying to create isn't working like I'd expect it to and I cant find the right words to type into the search to find them.

The interface so far is:

public <T extends SDOBase> T entityToSDO(<? extends BaseEntity> entity, T sdo) throws Exception;

How do I make entity a second generic type?

I have two separate types that need to go in there and they both extend different base classes. The second wildcard is incomplete as I couldn't figure out the syntax to express what I needed. — MikeG
– MikeG, Commented May 25, 2012 at 18:34
Oh nevermind, I just re-read your code and posted an answer. — newacct
– newacct, Commented May 25, 2012 at 22:03

Jeshurun · Accepted Answer · 2012-05-25 18:02:17Z

2

Try declaring your method this way:

public <T extends SDOBase, E extends BaseEntity> T entityToSDO(E entity, T sdo) throws Exception;

answered May 25, 2012 at 18:02

Jeshurun

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dnc253 · Accepted Answer · 2012-05-25 18:01:44Z

0

I think you want something more along the lines of:

public interface MyInterface<T extends SDOBase>
{
    public T entityToSDO(<? extends BaseEntity> entity, T sdo) throws Exception;
}

answered May 25, 2012 at 18:01

dnc253

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1 Comment

Paul Bellora Over a year ago

<? extends BaseEntity> entity isn't valid syntax for a method parameter.

newacct · Accepted Answer · 2012-05-25 22:02:24Z

0

You just need

public <T extends SDOBase> T entityToSDO(BaseEntity entity, T sdo) throws Exception;

Anything that is an instance of a subtype of BaseEntity automatically is an instance of BaseEntity.

answered May 25, 2012 at 22:02

newacct

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