Why do I get a compilation error when I try to have two methods with the same name and parameter type?

Question

If i change the byte to int I get a compiler error. Could you explain the problem?

public class A {
   protected int xy(int x) { return 0; }
}

class B extends A {
   protected long xy(int x) { return 0; } //this gives compilor error
   //protected long xy(byte x) { return 0; } // this works fine
}   

Answer 1

If i change the byte to int I get a compiler error.

If you do that, you have this:

public class A {
   protected int xy(int x) { return 0; }
}

class B extends A {
   protected long xy(int x) { return 0; }
}   

...and the only difference in the xy methods is their return type. Methods cannot be differentiated solely by their return types, that's the way Java is defined. Consider this:

myInstance.xy(1);

Which xy should that call? long xy(int x) or int xy(int x)?

If your goal is to override xy in B, then you need to make its return type int in order to match A#xy.