return type determines override/overload?

[hedefalk]

Hi!

This might really seem like a beginner's question and it probably is, but I want the Expert's answer

Consider the following Java 5-code:

Code:

class Request<T> {
	private T t;

	public T getT() {
		return t;
	}

	public void setT(T t) {
		this.t = t;
	}
	
}

class A {
	public Integer id(Request<Integer> request) {
		return new Integer(0);
	}
	
	public Double id(Request<Double> request) {
		return new Double(0.0);
	}
}

class B {
	public Double id(Request<Integer> request) {
		return new Double(0.0);
	}
	
	public Double id(Request<Double> request) {
		return new Double(0.0);
	}
}

In this code class A compiles just fine, but class B doesn't. You get an error saying there are duplicate methods. First off, I was hoping that since the type of a generic parameter is known at compile time, these methods in class B would be overloaded. I know that the generics is only a compile time construction, but I thought that this would work anyway, maybe by inserting this type information in the identifier or something.

This is not my big problem though. What I do not understand is why the same thing works just fine in class A when the only thing I have changed is the return type. In all functional and imperative programming languages I have ever been in contact with the signature of a function or method is always determined by its name and its parameter types. And I would wanna believe that it is the signature that determines if to methods clashes or not and the case of inheritance, if they are overridden or overloaded. But here, a difference in the return type changes this behaviour.

I would be grateful for an explaination!

/Viktor

[mary]

both method from B class do the same thing, they have the same parameters, check it:
you have this

Code:

class A {
public Double id(Request<Integer> request) {
return new Integer(0);
}

public Double id(Request<Double> request) {
return new Double(0.0);
}

and you have to do that:

Code:

class A {
public Integer id(Request<Integer> request) {
return new Integer(0);
}

public Double id(Request<Double> request) {
return new Double(0.0);
}

good luck

[hedefalk]

Yes, I understand that both methods do the same thing. I constructed them as simple as possible to focus on the point that the only difference between the to classes A and B is the return type of the methods.

[hedefalk]

I'll try to make an even simpler example to focus on the point of my question:

Code:

class Generic<T> {
}

class CompilesFine {
	public int id(Generic<Integer> request) {
		return 0;
	}
	
	public byte id(Generic<Double> request) {
		return 0;
	}
}

class DoesNotCompile {
	public int id(Generic<Integer> request) {
		return 0;
	}
	
	public int id(Generic<Double> request) {
		return 0;
	}
}

/Viktor

[mary]

I tested them, and you're right.
The only thing that I think it's maybe <> make them indistinguishable, and that is the reason about "duplicate method"