scjp

Please take a look at below example i cant understand the relation between char and byte byte b = 1; char c = 2; c = b; // line 1 Give me compilation Error because c is type of char and b is type of byte so casting is must in such condition but now the tweest here is when i run below code final byte b = 1; char c = 2; c = b; // line 2 line 2 compile successfully it doesn't need any casting at all so my question is why char c behave different when i use final access modifier with byte Because qualifying it with final makes the variable a constant variable, which is a constant expression . So

K.Sierra in her book "SCJP Study Guide" mentions "We know that a literal integer is always an int, but more importantly, the result of an expression involving anything int-sized or smaller is always an int." I've started experimenting and I'm a little bit confused with the below results: byte a = 1; // correct byte b = 1 + a; // incorrect (needs explicit casting) byte c = 1 + 1; // correct (I expected it to be incorrect) Could anyone explain to me why the last example does not require casting? Why does the Java compiler put the implicit cast? Is it because there are 2 int literals?

consider this sample code: 1. public class GC { 2. private Object o; 3. private void doSomethingElse(Object obj) { o = obj; } 4. public void doSomething() { 5. Object o = new Object(); 6. doSomethingElse(o); 7. o = new Object(); 8. doSomethingElse(null); 9. o = null; 10. } 11. } When the doSomething method is called, after which line does the Object created in line 5 become available for garbage collection? A. Line 5 B. Line 6 C. Line 7 D. Line 8 E. Line 9 F. Line 10 Answer: D why D? it's true that when Line 6 is executed the object created in Line 5 is now referenced by the instance var o and

I've got some doubts regarding protected identifier. In the first chapter of Sun Certified Java Programmer Study Guide by K.Sierra I found the following information: "Once the subclass-outside-the-package inherits the protected member, that member (as inherited by the subclass) becomes private to any code outside the subclass, with the exception of subclasses of the subclass." I provided sample code which reflects the above statement and it is absolutely clear to me. // Parent class package package1; import package2.Child; public class Parent { protected int i = 5; } // Child class package

Suppose a thread A is running. I have another thread, B , who's not. B has been started, is on runnable state. What happens if I call: B.join() ? Will it suspend the execution of A or will it wait for A's run() method to complete? brabster join() will make the currently executing thread to wait for the the thread it is called on to die. So - If A is running, and you call B.join(), A will stop executing until B ends/dies. Join waits till the thread is dead. If you call it on a dead thread, it should return immediately. Here's a demo: public class Foo extends Thread { /** * @param args */ public