Answer / rajmallela

Java is platform independent. but JVM is not platform independent .it is platform dependent.Each O.S has its own JVM
That JVM can understand .class file irrespective of the O.S from which it(.class file) is coming . So widows generated .class file will work in Linux O.S.(b/z Linux JVM can understand .class file obviously)

Answer / seshadri pera

Ranjmella is obsolutely correct for this Question. JVM is
platform dependent but any JVM specific to that OS will run
the compiled .class file on any OS.

.class files are in bytecode it can be run on any platform
which has it's specific JVM installed in that.

Answer / chandu

It will work, becouse java is platform independent.

Answer / stranger_ajnabi53@yahoo.c

ya it will work.bcoz java is platform independent.

Answer / anjan singh

Whenever you compile a java code, a bytecode is generated
(i.e. class file). This bytecode can be interpreted by JVM.
So any machine with any platform(i.e.OS) that has JVM on it
can interprete this bytecode or class file.

What is scope & storage allocation of global and extern variables? Explain with an example

0 Answers   IBS,

Question 7 [8] Consider the following class and answer the questions below it: public class StackWithGuard extends Stack { public StackWithGuard(int size) { super(size); } synchronized public boolean isEmpty() { return super.isEmpty(); } synchronized public boolean isFull() { return super.isFull(); } synchronized public int getSize() { return super.getSize(); } synchronized public void push(Object obj) { try { while (isFull()) { wait(); } } catch (InterruptedException e) {} super.push(obj); COS2144/102 11 notify(); } synchronized public Object pop() { try { while (isEmpty()) { wait(); } } catch (InterruptedException e) {} Object result = super.pop(); notify(); return result; } public static void main(String args[]) { StackWithGuard stack = new StackWithGuard(5); new Producer(stack, 15).start(); new Consumer(stack, 15).start(); } } Note: The Stack class is provided in the Appendix. Note also: The following questions all refer to the pop() method of the StackWithGuard class given above. 7.1 What does the synchronized keyword ensure for this method? (2) 7.2 Why is a while loop used to test whether the stack is empty? In other words, why wouldn't the following if statement be sufficient? if (isEmpty()) { wait(); } (2) 7.3 Why is the result of popping (provided by the inherited pop() method) stored in a temporary variable? In other words, why wouldn't the following statement be sufficient? return super.pop(); (2) 7.4 Why is the while loop placed in a try-catch structure? (2) Appendix The LinkedQueue class: public class LinkedQueue implements Queue { private Node first, last; private int count; public LinkedQueue() { first = last = null; count =0; } public int size() { return count; } public boolean isEmpty() { return (count == 0); 12 } public void enqueue(Object o) { Node node = new Node(); node.element = o; node.next = null; node.prev = last; if (last != null){ last.next = node; } else { last = first = node; } last = node; count++; } public void dequeue() { if ((first!= null) & (first.next!=null)) { first = first.next; first.prev = null; count--; } else { first = last = null; count--; } } public Object front() { return first; } } class Node { Object element; Node next, prev; } The Stack class: public class Stack { protected Object rep[]; protected int top = -1; protected int size = 0; protected int count = 0; public Stack(int size) { if (size > 0) { this.size = size; rep = new Object[size]; } } public boolean isFull() { return (count == size); } public boolean isEmpty() { return (count == 0); } public int getSize() { return size; } public void push(Object e) { if (e != null && !isFull()) { COS2144/102 13 top++; rep[top] = e; count ++; } } public Object pop() { Object result = null; if (!isEmpty()) { result = rep[top]; top--; count--; } return result; } }

0 Answers  

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0 Answers   Global Logic,

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0 Answers