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if no bound was specified. How does that work if a type parameter has several bounds?

April 26, 2017bound bounds parameter type

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if no bound was specified. How does that work if a type parameter has several bounds?

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Example (before type erasure): interface Runnable { void run(); } interface Callable { V call(); } class X & Runnable> { private T task1, task2; … public void do() { task1.run(); Long result = task2.call(); } } Example (after type erasure): interface Runnable { void run(); } interface Callable { Object call(); } class X { private Callable task1, task2; … public void do() { ( (Runnable) task1).run(); Long result = (Long) task2.call(); } } The type parameter T is replaced by the bound Callable , which means that both fields are held as references of type Callable . Methods of the leftmost bound (which is Callable in our example) can be called directly. For invocation of methods of the other bounds ( Runnable in our example) the compiler adds a cast to the respective bound type, so that the methods are accessible. The inserted cast cannot fail at runtime with a ClassCastException because the compiler already made sure at compile-time that both fields are re