Java Generics Hell

问题

I suspect this has been asked here (and answered) before, but I don't know how to name the problem. Why can I express the wildcards without problem only when I'm not passing the class itself?

It all boils down to this code. Everything works as expected except for the call to genericsHell(ShapeSaver.class):

interface Shape { }

interface Circle extends Shape { }

interface ShapeProcessor<T extends Shape> { }

class CircleDrawer implements ShapeProcessor<Circle> { } 

class ShapeSaver<T extends Shape> implements ShapeProcessor<T> { }

class Test {
    void genericsHeaven(ShapeProcessor<? extends Shape> a) {}

    void genericsHell(Class<? extends ShapeProcessor<? extends Shape>> a) {}

    void test() {
        genericsHeaven(new CircleDrawer());
        genericsHeaven(new ShapeSaver<Circle>());
        genericsHell(CircleDrawer.class);
        genericsHell(ShapeSaver.class); // ERROR: The method genericsHell is not applicable for the arguments (Class<ShapeSaver>)
    }
}

回答1:

The type of ShapeSaver.class is Class<ShapeSaver>. When feed it to genericsHell(), compiler needs to check if Class<ShapeSaver> is a subtype of Class<? extends ShapeProcessor<?>, which is reduces to whether ShapeSaver is a subtype of ShapeProcessor<?>. The subtype relation does not hold, the method call fails.

The same thing should be true for @Bohemian's solution. Here the subtype checking occurs at bound checking of T after T is inferred. It should fail too. This appears to be a compiler bug, which somehow misinterprets the rule that Raw is assignable to Raw<X> as if Raw is a subtype of Raw<X>. see also Enum.valueOf throws a warning for unknown type of class that extends Enum?

A simple solution to your problem is to declare

void genericsHell(Class<? extends ShapeProcessor> a)

indeed, ShapeSaver is a subtype of ShapeProcessor, and the call compiles.

That's not just a workaround. There's a good reason for it. Strictly speaking, for any Class<X>, X must be a raw type. For example, Class<List> is ok, Class<List<String>> is not. Because there is really no class that represents List<string>; there is only a class representing List.

Ignore the stern warning that you shall not use raw type. We must use raw types sometimes, given how Java type system is designed. Even Java's core APIs (Object.getClass()) use raw types.

---

You probably intended to do something like this

genericsHell(ShapeSaver<Circle>.class);

Unfortunately, that's not allowed. Java could have, but did not, introduce type literal along with generics. That created lots of problems for lots of libraries. java.lang.reflect.Type is a mess and unusable. Every library has to introduce their own representation of type system to solve the problem.

You can borrow one, e.g. from Guice, and you'll be able to

genericsHell( new TypeLiteral< ShapeSaver<Circle> >(){} )
                               ------------------  

(learn to skip the craps around ShaveSaver<Circle> when reading the code)

In the method body of genericsHell(), you'll have full type information, not just the class.

回答2:

Typing the genericsHell method allows it to compile:

static <T extends ShapeProcessor<?>> void genericsHell(Class<T> a) {}

EDITED: This allows the compiler to specify from context, or by coding an explicit type, that the ShapeProcessor is not literally any ShapeProcessor, but the same type as the one passed as a parameter. If the call was explicitly typed, (which the compiler does under the covers) the code would look like this:

MyClass.<ShapeSaver>genericsHell(ShapeSaver.class);

Which interestingly, gives a type warning, but still compiles. The explicit type is not required however because sufficient type information is available from the parameter to infer the generic type.

Your question is missing some declarations, so I added them in to create a Short Self-Contained Correct Example - ie this code compiles as-is

static interface Shape { }

static interface Circle extends Shape { }

static interface ShapeProcessor<T extends Shape> { }

static class CircleDrawer implements ShapeProcessor<Circle> { }

static class ShapeSaver<T extends Shape> implements ShapeProcessor<T> { }

static void genericsHeaven(ShapeProcessor<? extends Shape> a) { }

// The change was made to this method signature:
static <T extends ShapeProcessor<?>> void genericsHell(Class<T> a) { }

static void test() {
    genericsHeaven(new CircleDrawer());
    genericsHeaven(new ShapeSaver<Circle>());
    genericsHell(CircleDrawer.class);
    genericsHell(ShapeSaver.class);
}

来源：https://stackoverflow.com/questions/7054842/java-generics-hell