Generics ambiguity with the &-operator and order

问题

I have a strange Java generics ambiguity behaviour that I cannot explain:

Those 3 methods in class:

public static <E extends ClassA & ClassB> void method(E val) {}
public static <E extends ClassC & ClassB & ClassA> void method(E val) {}
public static <E extends ClassB> void method(E val) {}

compile fine.

But those not (ambiguity violation):

public static <E extends ClassA & ClassB> void method(E val) {}
public static <E extends ClassB & ClassC & ClassA> void method(E val) {}
public static <E extends ClassB> void method(E val) {}

(ClassA, ClassB, ClassC are all completely independent interfaces!)

回答1:

Due to type erasure, the compiler needs to pick a statically-known type for the parameter type in the compiled method.

To do this, it uses the first type in your constraint list.

In your first example, this results in a unique type for each method, so it compiles to

public static method(ClassA val);
public static method(ClassC val);
public static method(ClassB val);

This is perfectly legal (except for your missing return type); it creates three overloads with three different parameter types.

In your second example, this creates an ambiguity:

public static method(ClassA val);
public static method(ClassB val);
public static method(ClassB val);

This is not legal, because the last two methods have the same signature.

The spec explicitly documents this behavior.

This could have been made legal by trying to pick a single constraint type from each overload such that there are no conflicts, but that would be complicated & slow for larger constraint lists.
The spec could have said something like:

If it is used in erasure of a type in the parameter list, the erasure of a type variable in a generic method is chosen such that each overload of that method results in a unique signature after erasure.
If no combination of erasures will result in a unique signature, an ambiguity error occurs.

I suspect that this problem is in NP.

回答2:

It is defined in the JLS #4.6:

The erasure of a type variable is the erasure of its leftmost bound.

And if two methods have the same erasure, the compiler gives you an error.

来源：https://stackoverflow.com/questions/21142467/generics-ambiguity-with-the-operator-and-order