About Question enthuware.ocpjp.v7.2.1308 :

[Wisevolk]

Hi,
class MidiPlayer implements MusicPlayer<Instrument> {     public void play(Guitar g){ } }
Your answer : MidiPlayer must have a method play(Object ).

Is it right ? Shouldn't Midi player have a a method play(Instruments) ?

[admin]

Yes, it is correct because the method is actually coming from interface Player<E>{ void play(E e); }

HTH,
Paul.

[lunars]

This is my current understanding of the problem, further dissecting the uncertainty brought up by Wisevolk (see comments):

Code: Select all

class Game{ } 
class Cricket extends Game{ } 
class Instrument{ } 
class Guitar extends Instrument{ }  
interface Player<E>{ void play(E e); } 
interface GamePlayer<E extends Game> extends Player<E>{ } 
interface MusicPlayer<E extends Instrument> extends Player{ } 
/** 
 *  (1) When MusicPlayer extends the *RAW* "Player" above, that makes 
 *       "void play(E e);" effectively convert into "void play(Object o);"
 *  (2) So anything implementing MusicPlayer (regardless of generics used) 
 *       gets handed a version of Player's method(s) where E (only the E 
 *       inside Player) has been downgraded to Object (due to being raw)
 */

class MidiPlayer implements MusicPlayer<Instrument> {
	
	public void play(Instrument i){ }  // <-- 
	public void play(Guitar g){ }      // <-- neither of these can reference the raw version of Player 
	
	public void play(Object e) {}      // <-- comes from raw version of Player
	
}

Is that accurate?

[admin]

That is correct, lunars.
-Paul.

[jagoneye]

This is my current understanding of the problem, further dissecting the uncertainty brought up by Wisevolk (see comments):

Code: Select all

class Game{ } 
class Cricket extends Game{ } 
class Instrument{ } 
class Guitar extends Instrument{ }  
interface Player<E>{ void play(E e); } 
interface GamePlayer<E extends Game> extends Player<E>{ } 
interface MusicPlayer<E extends Instrument> extends Player{ } 
/** 
 *  (1) When MusicPlayer extends the *RAW* "Player" above, that makes 
 *       "void play(E e);" effectively convert into "void play(Object o);"
 *  (2) So anything implementing MusicPlayer (regardless of generics used) 
 *       gets handed a version of Player's method(s) where E (only the E 
 *       inside Player) has been downgraded to Object (due to being raw)
 */

class MidiPlayer implements MusicPlayer<Instrument> {
	
	public void play(Instrument i){ }  // <-- 
	public void play(Guitar g){ }      // <-- neither of these can reference the raw version of Player 
	
	public void play(Object e) {}      // <-- comes from raw version of Player
	
}

Is that accurate?

Did not understand this and also why
MidiPlayer must have a method play(Object ).
Why can't I use type Instrument here?

[admin]

Try compiling using Instrument and see what the compiler error message says. That should help.

[jagoneye]

Actually I was trying to verify if this would work

Code: Select all

class MidiPlayer implements MusicPlayer<Instrument>
{
    public void play(Instrument o){ } 
}

And I didn't see that MusicPlayer was extending RAW Player!!!
What I thought was

Code: Select all

interface MusicPlayer<E extends Instrument> extends Player<E>{ }

and if this was the interface then my MidiPlayer will happily compile!