I have would like to create an interface:
public interface OperandValue { <T> T getValue(); }
I would then like to have a concrete implementation like this:
public class NumberOperandValue implements OperandValue { @Override public <Integer> Integer getValue() { // some integer value that is set elsewhere return 1; } }
Eclipse is underlining the <Integer> giving me a warning that says:
The type parameter Integer is hiding the type Integer
I appreciate any suggestions on if this can work somehow. I realize that I could define the generic type at the interface level and not the method level, but would like to try to get this to work if possible.
You probably want to change the interface to:
public interface OperandValue<T> { T getValue(); }
And the implementation to:
public class NumberOperandValue implements OperandValue<Integer> { @Override public Integer getValue() { // some integer value that is set elsewhere return 1; } }
Now you're telling the interface what type you want that method to return. In other words, you're making the interface type generic, rather than the method declaration. But that seems to be what you want.
As a side note:
public <Integer> Integer getValue()
Actually means, 'define a generic type parameter with the name "Integer"' where getValue returns the "Integer" type that you just defined.
Responding To Steve's Comments Below:
when I remove the <Integer> from my method implementation that I then get a warning that says: Type safety: The return type Integer for getValue() from the type NumberOperandValue needs unchecked conversion to conform to T from the type OperandValue
That warning message indicates that you're breaking the rules when using Java generics. To see why, let's consider what the method signature means before you remove the <Integer> type parameter.
public <Integer> Integer getValue()
This signature means that the method getValue returns a value of type Integer where Integer is defined as the generic type parameter you defined between the angle brackets. The meaning of the string Integer is completely arbitrary and would have exactly the same meaning as:
public <T> T getValue()
For clarity, let's stick with this version of your method signature for the purposes of your question. What happens when we remove the type parameter?
public T getValue()
Now if you were to try to compile, you'd get an error that T is undefined. However, because your original type signature declared the type parameter with the name Integer, when you removed it, you were left with:
public Integer getValue()
Because Integer is already a predefined type, the method signature is still technically legal. However, it is only an accident that the name of your type parameter happens to be the same as a type that already exists.
Furthermore, because your interface already declared a method signature with generics, the Java compiler generates a warning when you remove it from the implementation. Specifically, the compiler is concerned that in the base class, the return type of the method is the (type-erased to Object) generic parameter named Integer, which is not the same type (nor is it known to be type compatible with) the system class named Integer (or java.lang.Integer to be precise).
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