Tag: programming languages

What is the output for the below code ? public class A { public A() { System.out.println("A"); } } public class B extends A implements Serializable { public B() { System.out.println("B"); } } public class Test { public static void main(String... args) throws Exception { B b = new B(); ObjectOutputStream save = new ObjectOutputStream(new FileOutputStream("datafile")); save.writeObject(b); save.flush(); ObjectInputStream restore = new ObjectInputStream(new FileInputStream("datafile")); B z = (B) restore.readObject(); } }

Please dont Try this Question... Try Next One.... If you attend others, you will get to know the answer of this question. Questions: 1. 1. public class Question { Public static void main(String args[]) { int i=10; System.out.println("i="+++i); } 2. public class Question { public static void main(String args[]) { int x=010; System.out.print("X="+x); } } 3. public class Question { public static void main(String args[]) { int x=y=10; y=12; System.out.print(x+"&"+y); } } 4. public class Question{ public static void main(String args[]) { int i=3,j=2; System.out.print("i|j="+ (i|j)); } } 5. public class Question { public static void main(String args[]) { int x=0x10; System.out.print("X="+x); } } Answers: 1. System.out.println("i="+++i); Must be as a System.out.println("i="+ ++i); “Space between operators” 2. int x=010; In this statement, compile will take it as a OCTAL NUMBER. So OCTAL(10)=DECIMAL(8) 3. int x=y=10; In this statement, We are using y without declaring it. So will cause Compile Time Error. 4. 3 ? 1 1 2 ? 1 0 ------------------ (OR) 3 ? 1 1 5. int x=0x10; In the above statement, Compile will take it as a HEXA DECIMAL. So. HEXA(10) ? DECIMAL(16)