Uses of T extends U?

Question

Lately I encountered a method defined similar to this and I don't exactly understand its usage:

public static <T, U extends T> T foo(U u) { ... }

A sample use could be like this:

// Baz is just the containing class of foo()
Number n = Baz.foo(1);

Where T is inferred to Number and U (probably) to Integer. But I can't wrap my head around when this is superior to e.g. this method definition:

public static <T> T bar(T t) { ... }

If I call it like this:

Number n = Baz.bar(2);

The code still works. T is inferred to either Number or Integer (Don't know if the argument type, in this example Integer is preferred over the call site return type Number)

I've read these questions: 1, 2 but I still don't know if the first method with 2 parameters has any advantage over the second method with only one generic.

Answer 1

I think that, in fact, this only makes sense when the type parameter of the method appears as the type parameter of a parameterized type that is part of the method signature.

^{(At least, I couldn't quickly come up with an example where it would really makes sense otherwise)}

This is also the case in the question that you linked to, where the method type parameters are used as type parameters in the AutoBean class.

_{A small update:}

Based on the discussion in the question and other answers, the core of this question was likely a misinterpretation of the way of how the type parameters are used. As such, this question could be considered as a duplicate of Meaning of <T, U extends T> in java function declaration , but hopefully someone will consider this answer helpful nevertheless.

In the end, the reason for using the pattern of <T, U extends T> can be seen in the inheritance relationships of parameterized types, which in detail may be fairly complicated. As an example, to illustrate the most relevant point: A List<Integer> is not a subtype of a List<Number>.

An example showing where it can make a difference is below. It contains a "trivial" implementation that always works (and does not make sense, as far as I can tell). But the type bound becomes relevant when the type parameters T and U are also the type parameters of the method parameters and return type. Whith the T extends U, you can return a type that has a supertype as the type parameter. Otherwise, you couldn't, as shown with the example that // Does not work:

import java.util.ArrayList;
import java.util.List;

public class SupertypeMethod {
    public static void main(String[] args) {

        Integer integer = null;
        Number number = null;

        List<Number> numberList = null;
        List<Integer> integerList = null;

        // Always works:
        integer = fooTrivial(integer);
        number = fooTrivial(number);
        number = fooTrivial(integer);

        numberList = withList(numberList);
        //numberList = withList(integerList); // Does not work

        // Both work:
        numberList = withListAndBound(numberList);
        numberList = withListAndBound(integerList);
    }

    public static <T, U extends T> T fooTrivial(U u) {
        return u;
    }

    public static <T, U extends T> List<T> withListAndBound(List<U> u) {
        List<T> result = new ArrayList<T>();
        result.add(u.get(0));
        return result;
    }

    public static <T> List<T> withList(List<T> u) {
        List<T> result = new ArrayList<T>();
        result.add(u.get(0));
        return result;
    }

}

(This looks a bit contrived, of course, but I think that one could imagine scenarios where this actually makes sense)