为什么不能使用签名相同的静态和非静态方法呢？

Question

I just realized the following does not compile:

class MyCoolClass {
    static void do(){..}
    void do(){..}
}

Is there a good reason why this is not allowed? Maybe it would cause some problem that I'm currently missing?

The only problem I can see currently is that someone might try to call the static method and actually calls the non-static one. But that sounds comparable with static methods on subclasses that hide each other, so I wouldn't consider that a "good" reason.

If one considers the object a method is called on the 0th argument to that method it even would be quite normal overloading of the method.

In case you are wondering: I would like to have this because it would allow creating a shortcut for new MyCoolClass().do() as MyCoolClass.do() or with static import even as do() .

Answer 1

It's because the Java compiler uses the name of the method and the ordered list of parameter types to uniquely identify the method within the class. static is a modifier flag and is not considered part of the unique identification. Unique identification is also important at runtime because the class file contains only those keys necessary to identify the method. The class name where the method is, then the method name, then the parameter list.

https://docs.oracle.com/javase/specs/jvms/se7/html/jvms-5.html#jvms-5.4.3.3

Answer 2

I think this might be problematic to handle when you are creating an object. As you probably know, you can call a static function through object reference like:

SomeClass s = new SomeClass();
s.do(); //can be static method, can be normal method

Of course, it's rare to call static method through object, but it's possible.