13.4.23

Method Overriding

BINARY COMPATIBILITY

class Super {

static void out(float f) { System.out.println("float"); }

static void out(int i) { System.out.println("int"); }

}

If

Super

 is recompiled but not

Test

, then running the new binary with the exist 

ing binary of

Test

 still produces the output:

float

However, if

Test

 is then recompiled, using this new

Super

, the output is then:

int

as might have been naively expected in the previous case.

13.4.23   Method Overriding

If an instance method is added to a subclass and it overrides a method in a super 

class, then the subclass method will be found by method invocations in pre exist 

ing binaries, and these binaries are not impacted. If a class method is added to a

class, then this method will not be found, because the invocation of a class method

is resolved at compile time to use the fully qualified name of the class where the

method is declared. Thus if the example:

class Hyper {

void hello() { System.out.print("Hello, "); }

static void world() { System.out.println("world!"); }

}

class Super extends Hyper { }

class Test {

public static void main(String[] args) {

Super s = new Super();

s.hello();

s.world();

}

}

is compiled and executed, it produces the output:

Hello, world!

Suppose that a new version of class

Super

 is produced:

class Super extends Hyper {

void hello() { System.out.print("Goodbye, cruel "); }

static void world() { System.out.println("earth!"); }

}

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