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Q52. If B extends A and you create object of B where default constructor of A throws some exception then whether object of B would be created? Ans. As per Inheritance rule, child class needs to call super() on first line of it's constructor. But as per Exception rule, it needs to be handled or thrown. So child class constructor will have to throw it again. See. If there is not runtime exception then the object will be created.

Q53. If C extends B extends A and you create C object by its parameterized constructor then whether constructor of A and B would be called? Ans. As per Inheritance rule, yes.

Q54. Is this scenario is possible? B extends A where A and B both have a parameterized constructor. Ans. As per Inheritance rule, child class need to call super class constructor otherwise compiler adds empty super() call. Since parent class have parameterized constructor, this call will fails. However if child can call parent parameterized constructor to make it successful.

Q55. Whether super.super or this.this is allowed? Ans. this.this is senseless. this.super is not allowed. super.super is not allowed because son inherits properties of father only while the father inherits properties of his father. So if C extends B extends A then B can access all visible members of A. Thus C can access members of A through super only. It doesn’t require super.super.

Q56. Can I have a final member field in a class which is not initialized anywhere in the class? Ans. As per Final rules, No. It gives compilation error.

Q57. A method() accepts String str as parameter then what will str.equals(null) output if I call method(null)? Ans. When you set str=null, it free the object it is referring to. So calling any method on this will cause NullPointerException.

Q58. If

String s1 = “Amit”;
String s2 = new String(“Amit”);

Which is true and why

  1. “Amit”.equals(s1);
  2. “Amit” == s1
  3. “Amit”.equals(s2);
  4. “Amit” == s2
  5. s1.equals(s2);
  6. s1 == s2

Ans.

  • As per Final rule, s1 will point to memory location where “Amit” resides. So 2nd statements is true. But 4th is false.
  • As per String rule, 1,3,5 are true. and 6 is false.

Exception Handling

Q59. If

try{
	String n = null;
	System.out.println(n.equals(null));
	System.out.println(1);
}catch(Exception e){
	System.out.println(2);
}catch(NullPointerException e){
	System.out.println(3);
}

What would be the output?

Ans. As per Exception rule, only 1 catch block can be called. Since Exception is parent of all type of exceptions. It'll be always called. Hence compile time error "Unreachable catch block for NullPointerException".

Q60. If

public int myMethod() {
	int i=0;
	try{
		System.out.println(i);
		return 2;
	}catch(Exception e){
		System.out.println(i);
		return 3;
	}finally{
		System.out.println(i);
		return 4;
	}
}
:
System.out.println("->" +s.myMethod());

What would be the output of above code? Ans. Above code will output

0
0
4

Note:

  1. Since finally block runs ever whether any exception occurs or not so return statements inside try & catch will be ignored. But their expression will be executed (like in above example ++i & i++).
    Since, in below program, finally returns program control to parent caller and return in try block doesn’t get completed so it gives abnormal completion warning.
  2. Below code can give Unreachable code compilation error if there is any statement after finally block.

Q61. What will be the output of below program?

public static int method(){
	int i = 1;
	try{
		i++;
		return ++i;
	}catch(Exception e){
		i++;
		return ++i;
	}finally{
		i++;
		System.out.println(i);
		return ++i;
	}
}

System.out.println(method());

Ans.

4
5

Q62. What will be the output of below program?

public static int method(){
	int i = 1;
	try{
		i++;
		return ++i;
	}catch(Exception e){
		i++;
		return ++i;
	}finally{
		i++;
		System.out.println(i);
	}
}

System.out.println(method());

Ans.

4
3

Serialization

public class GrandParent {
	public GrandParent() {
		System.out.println("Grand Parent");
	}
}

public class Parent extends GrandParent implements Serializable{
	public Parent() {
		System.out.println("Parent");
	}
}

public class Child extends Parent{
	public Child() {
		System.out.println("Child");
	}
	public static void main(String[] args) throws Exception {
		Child child = new Child();
		
		ObjectOutputStream out = new ObjectOutputStream(new FileOutputStream("fileName"));
		out.writeObject(child);
		
		System.out.println("Deserialing");
		
		ObjectInputStream in = new ObjectInputStream(new FileInputStream("fileName"));
		Child child2 = (Child) in.readObject();
	}
}

Output

Grand Parent
Parent
Child
Deserialing
Grand Parent

Q63. If C extends B extends A where

  1. A is serializable
  2. B is serializable
  3. A and B both are serializable
  4. A and B both are non-serializable

Which constructors will be called by Deserialization?

Ans. As per serialization rule 3,

  1. No constructor call. Because deserialization finds serializable class in starting
  2. A()
  3. No constructor call
  4. A() then B()

Q64. If C extends B extends A where

  1. A is serializable
  2. B is serializable
  3. A and B both are serializable
  4. A and B both are non-serializable

Where A has no default constructor but parameterized constructor().

Which constructors will be called by Deserialization? Ans.

  1. No constructor call. Because deserialization finds serializable class in starting
  2. It’ll give compilation error since, as per Inheritance rule, parent class always must have default constructor or child class constructor must have explicit call to parameterized constructor of parent class.
    However if child class call parameterized constructor of super class then, it'll throw runtime exception at the time of deserialization: "java.io.InvalidClassException".
  3. No constructor call
  4. Same as point 2

Enum

Q65. Why an enum can not extend another enum or class but an interface while the java compiler translates it into class later?
And. As per Enum rule and inheritance, it already extends Enum. That class provides all the enum functionality.

Q66. Whether an enum can have abstract methods? Ans. Yes. Since all the instances declared inside the enum can have body of anonymous class. So they have to override abstract. Check this.

Q . Should I use class or Enum for Directions Ans. Considering, directions has fixed 4 types: North, East, West, South. There is no point to create multiple instances of any of directions. So all the Directions types are singleton. Hence, Using enum would be right choice.

Generics

Q67. Explain below code

List<Fruit> fruits = new ArrayList<Fruit>();
fruits.add(new Apple());
fruits.add(new Strawberry());

List<Fruit> fruits = new ArrayList<Apple>();//Compilation Error: not allowed

Ans. As per Generics rule, ChildClass< B> is not the subtype of ParentClass< A>. Hence List< Apple> is not sub-type of List.

But a List< A> can contains all elements of type A so Apple,Strawberry are allowed to go in Fruit list.

Q68. Explain below code

Apple[] apples = new Apple[1];
Fruit[] fruits = apples;
fruits[0] = new Strawberry();//Runtime Exception

List<Apple> apples = new ArrayList<Apple>();
//List<Fruit> fruits = apples; //Compile time error
List<? extends Fruit> fruits = apples;
//fruits.add(new Strawberry()); //Compile time error
//fruits.add(new Fruit()); //Compile time error
//fruits.add(new Apple()); //Compile time error
Fruit fruit = fruits.get(0);


List<? super Apple> apples = new ArrayList<Fruit>();
apples.add(new Strawberry()); //Compile time error
apples.add(new Fruit()); //Compile time error
apples.add(new Apple());

Ans. For

Apple[] apples = new Apple[1];
Fruit[] fruits = apples;
fruits[0] = new Strawberry();//Runtime Exception

fruites is a reference pointer (say A) to an array (type of B). Since B extends A, A can be used as reference to B[]. However B[] can accept an element of type B only.

Now in case of generics, by the rule;

List<Apple> apples = new ArrayList<Apple>();
//List<Fruit> fruits = apples; //Compile time error

ChildClass< B> is not the subtype of ParentClass< A>. Hence it gives compilation error. Next to it;

List<? extends Fruit> fruits = apples;
//fruits.add(new Strawberry()); //Compile time error
//fruits.add(new Fruit()); //Compile time error
//fruits.add(new Apple()); //Compile time error
Fruit fruit = fruits.get(0);

List<? extends A> a can point to any list< B>, list etc. if B, C,.. are the type of A. But in this case, since you are not sure what a is pointing to. So if it is pointing to list< B> type then only items of type B can be added. Hence aadition to such list is not allowed. However Fruit fruit = fruits.get(0) will always return an item os type fruit only, it is allwed.

In last,

List<? super Apple> apples = new ArrayList<Fruit>();
apples.add(new Strawberry()); //Compile time error
apples.add(new Fruit()); //Compile time error
apples.add(new Apple());


In case of super, list of fruits or list of objects can be assigned to list of apples.
But while reading, you can't guarantee what will come out. However it is guaranteed to be Object.
While adding, since list of apples can point either list of apples itself or the list of it's super classes so any object of Apple or it's subtype can be added. But instance of Fruit or it's other type can't be added.

Q69. What is the following class converted to after type erasure?

public class Pair<K, V> {

    public Pair(K key, V value) {
        this.key = key;
        this.value = value;
    }

    public K getKey(); { return key; }
    public V getValue(); { return value; }

    public void setKey(K key)     { this.key = key; }
    public void setValue(V value) { this.value = value; }

    private K key;
    private V value;
}

Ans.

public class Pair {

    public Pair(Object key, Object value) {
        this.key = key;
        this.value = value;
    }

    public Object getKey()   { return key; }
    public Object getValue() { return value; }

    public void setKey(Object key)     { this.key = key; }
    public void setValue(Object value) { this.value = value; }

    private Object key;
    private Object value;
}

Q70. What is the following method converted to after type erasure?

public static <T extends Comparable<T>>
    int findFirstGreaterThan(T[] at, T elem) {
    // ...
}

Ans.

public static int findFirstGreaterThan(Comparable[] at, Comparable elem) {
    // ...
}

Q71. Justify below code snippet

SomeClass<?> someObj = new SomeClass<B>();
//but
someObj.someMethod(new A());//not allowed
someObj.someMethod(new B());//not allowed
someObj.someMethod(null);

Ans. ? represents no type. So A,B are definetely not type of ?. Hence only null is allowed.

Q72. What is the advantage of code given in Q69? Ans. If we write the above code in following way;

SomeClass<B> someBObj = new SomeClass<B>();
someBObj.someMethod(new B());
someBObj.someMethod(new B());
:
SomeClass<?> readOnlyList = someBObj;
B b = (B) readOnlyList.get(0);

Q73 Will the following class compile? If not, why?

public final class Algorithm {
    public static <T> T max(T x, T y) {
        return x > y ? x : y;
    }
}

Ans. No. The greater than (>) operator applies only to primitive numeric types.

Q74: Does following code is valid?

T t = new T();

//or
class Foo<T> {
 public void doSomething(T t) {
 	t.sort();
 }
}

Ans. As T is unkown, above code is not allowed. However following code can be used instead;

class Foo<T extends SomeClass> {
 public void doSomething(T t) {
 	t.sort();//  sort method presents in SomeClass
 }
}

NIO

Q75: Why FileChannel can't be use with Selector?
Ans: As FileChannel is blocking, by the rules, it can't be used with Selectors.

Q76: Does NIO are always faster than IO?
Ans: As NIO use less threads, it saves time and memory in thread(context) switching. Even the CPU utilization is less. But performance completely depends on what way code is written and what is being performed. In general NIO are faster.