26. Which of the following lines will print false?
1.public class MyClass
2.{
3.static String s1 = "I am unique!";
4.public static void main(String args[])
5.{
6.String s2 = "I am unique!";
7.String s3 = new String(s1);
8.System.out.println(s1 == s2);
9.System.out.println(s1.equals(s2));
10.System.out.println(s3 == s1);
11.System.out.println(s3.equals(s1));
12.System.out.println(TestClass.s4 == s1);
13.}
14.}
15.
16.class TestClass
17.{
18.static String s4 = "I am unique!";
19.}
A. line 10 and 12
B. line 12 only
C. line 8 and 10
D. none of these
D is correct. Only line 10 will print false. Strings are immutable objects. That is, a string is read only once the string has been created and initialized, and Java optimizes handling of string literals; only one anonymous string object is shared by all string literals with the same contents. Hence in the above code the strings s1, s2 and s4 refer to the same anonymous string object, initialized with the character string: "I am unique!". Thus s1 == s2 and TestClass.s4 will both return true and obviously s1.equals(s2) will return true. But creating string objects using the constructor String(String s) creates a new string, hence s3 == s1 will return false even though s3.equals(s1) will return true because s1 and s3 are referring to two different string objects whose contents are same.
27. What is displayed when the following code is compiled and executed?
String s1 = new String("Test");
String s2 = new String("Test");
if (s1==s2) System.out.println("Same");
if (s1.equals(s2)) System.out.println("Equals");
A. same equal
B. equals
C. same
D. compile but nothing is displayed upon exception
E. the code fails to compile.
B is correct. Here s1 and s2 are two different object references, referring to different objects in memory. Please note that operator == checks for the memory address of two object references being compared and not their value. The "equals()" method of String class compares the values of two Strings. Thus s1==s2 will return "false" while s1.equals(s2) will return "true". Thus only "Equals" will be printed.
28. What is displayed when the following is executed?
class Parent{
private void method1(){
System.out.println("Parent's method1()");
}
public void method2(){
System.out.println("Parent's method2()");
method1();
}
}
class Child extends Parent{
public void method1(){
System.out.println("Child's method1()");
}
public static void main(String args[]){
Parent p = new Child();
p.method2();
}
}
A. compile time error
B. run time error
C. prints: parent’s method2() parent’s method1()
D. prints: parent’s method2() child’s method1()
C is correct. The code will compile without any error and also will not give any run time error. The variable p refers to the Child class object. The statement p.method2() on execution will first look for method2() in Child class. Since there is no method2() in child class, the method2() of Parent class will be invoked and thus "Parent's method2()" will be printed. Now from the method2() , there is a call to method1(). Please note that method1() of Parent class is private, because of which the same method (method1() of Parent class) will be invoked. Had this method(method1() of Parent class) been public/protected/friendly (default), Child's class method1() would be called. Thus C is correct answer.
Given class parent method1 be public, while class child method1 be pivate, there will be a compile error, for method1 in class child tries to give a weak overridden; given class parent method1 be private, and main() in class child used p.method1(), there will be another compile error for method1 in class parent is not accessible; given class parent method1 be public and class child method1 be public, the end will be “prints: parent’s method2() child’s method1()”; given method1 in class child be private, there will be a compile error which is “mehod1 in class child is not accessible”.
29. What will happen when you attempt to compile and run the following code snippet?
String str = "Java";
StringBuffer buffer = new StringBuffer(str);
if(str.equals(buffer)){
System.out.println("Both are equal");
}else{
System.out.println("Both are not equal");
}
A. it will print – both are not equal
B. it will print – both are equal
C. compile time error
D. Runtime error
A is the correct choice. The equals method overridden in String class returns true if and only if the argument is not null and is a String object that represents the same sequence of characters as this String object. Hence, though the contents of both str and buffer contain "Java", the str.equals(buffer) call results in false.
The equals method of Object class is of form -public boolean equals(Object anObject). Hence, comparing objects of different classes will never result in compile time or runtime error.
30. What will happen when you attempt to compile and run the following code?
public class MyThread extends Thread{
String myName;
MyThread(String name){
myName = name;
}
public void run(){
for(int i=0; i<100;i++){
System.out.println(myName);
}
}
public static void main(String args[]){
try{
MyThread mt1 = new MyThread("mt1");
MyThread mt2 = new MyThread("mt2");
mt1.start();
// XXX
mt2.start();
}catch(InterruptedException ex){}
}
}
A. compile error
B. mt1.join();
C. mt1.sleep(100);
D. mt1.run()
E. nothing need
Choice A and B are correct. In its current condition, the above code will not compile as "InterruptedException" is never thrown in the try block. The compiler will give following exception: "Exception java.lang.InterruptedException is never thrown in the body of the corresponding try statement."
Note that calling start() method on a Thread doesn't start the Thread. It only makes a Thread ready to be called. Depending on the operating system and other running threads, the thread on which start is called will get executed. After making the above code to compile (by changing the InterruptedException to some other type like Exception), the output can't be predicted (the order in which mt1 and mt2 will be printed can't be guaranteed). In order to make the MyThread class prints "mt1" (100 times) followed by "mt2" (100 times), mt1.join() can be placed at //XXX position. The join() method waits for the Thread on which it is called to die. Thus on calling join() on mt1, it is assured that mt2 will not be executed before mt1 is completed. Also note that the join() method throws InterruptedException, which will cause the above program to compile successfully. Thus choice A and B are correct.
