What will be the output of the program?
class Super
{
public Integer getLength()
{
return new Integer(4);
}
}
public class Sub extends Super
{
public Long getLength()
{
return new Long(5);
}
public static void main(String[] args)
{
Super sooper = new Super();
Sub sub = new Sub();
System.out.println(
sooper.getLength().toString() + "," + sub.getLength().toString() );
}
}
Answer: D
Option D is correct, compilation fails - The return type of getLength( ) in the super class is an object of reference type Integer and the return type in the sub class is an object of reference type Long. In other words, it is not an override because of the change in the return type and it is also not an overload because the argument list has not changed.
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What will be the output of the program?
public class Test
{
public int aMethod()
{
static int i = 0;
i++;
return i;
}
public static void main(String args[])
{
Test test = new Test();
test.aMethod();
int j = test.aMethod();
System.out.println(j);
}
}
Answer: D
Compilation failed because static was an illegal start of expression - method variables do not have a modifier (they are always considered local).
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Given a method in a protected class, what access modifier do you use to restrict access to that method to only the other members of the same class?
Answer: C
The private access modifier limits access to members of the same class.
Option A, B, D, and E are wrong because protected are the wrong access modifiers, and final, static, and volatile are modifiers but not access modifiers.
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What will be the output of the program?
import java.util.*;
public class NewTreeSet2 extends NewTreeSet
{
public static void main(String [] args)
{
NewTreeSet2 t = new NewTreeSet2();
t.count();
}
}
protected class NewTreeSet
{
void count()
{
for (int x = 0; x < 7>
Answer: D
Nonnested classes cannot be marked protected (or final for that matter), so the compiler will fail at protected class NewTreeSet.
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What is the widest valid returnType for methodA in line 3?
public class ReturnIt
{
returnType methodA(byte x, double y) /* Line 3 */
{
return (long)x / y * 2;
}
}
Answer: A
However A, B and C are all wrong. Each of these would result in a narrowing conversion. Whereas we want a widening conversion, therefore the only correct answer is D. Don't be put off by the long cast, this applies only to the variable x and not the rest of the expression. It is the variable y (of type double) that forces the widening conversion to double.
Java's widening conversions are:
- From a byte to a short, an int, a long, a float, or a double.
- From a short, an int, a long, a float, or a double.
- From a char to an int, a long, a float, or a double.
- From an int to a long, a float, or a double.
- From a long to a float, or a double.
- From a float to a double.
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Which two of the following are legal declarations for nonnested classes and interfaces?
1.final abstract class Test {}
2.public static interface Test {}
3.final public class Test {}
4.protected abstract class Test {}
5.protected interface Test {}
6.abstract public class Test {}
Answer: C
(3), (6). Both are legal class declarations.
(1) is wrong because a class cannot be abstract and final—there would be no way to use such a class. (2) is wrong because interfaces and classes cannot be marked as static. (4) and (5) are wrong because classes and interfaces cannot be marked as protected.
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What will be the output of the program?
class Super
{
public int i = 0;
public Super(String text) /* Line 4 */
{
i = 1;
}
}
class Sub extends Super
{
public Sub(String text)
{
i = 2;
}
public static void main(String args[])
{
Sub sub = new Sub("Hello");
System.out.println(sub.i);
}
}
Answer: D
A default no-args constructor is not created because there is a constructor supplied that has an argument, line 4. Therefore the sub-class constructor must explicitly make a call to the super class constructor:
public Sub(String text)
{
super(text); // this must be the first line constructor
i = 2;
}
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What will be the output of the program?
interface Count
{
short counter = 0;
void countUp();
}
public class TestCount implements Count
{
public static void main(String [] args)
{
TestCount t = new TestCount();
t.countUp();
}
public void countUp()
{
for (int x = 6; x>counter; x--, ++counter) /* Line 14 */
{
System.out.print(" " + counter);
}
}
}
Answer: E
The code will not compile because the variable counter is an interface variable that is by default final static. The compiler will complain at line 14 when the code attempts to increment counter.
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What will be the output of the program?
class Base
{
Base()
{
System.out.print("Base");
}
}
public class Alpha extends Base
{
public static void main(String[] args)
{
new Alpha(); /* Line 12 */
new Base(); /* Line 13 */
}
}
Answer: B
Option B is correct. It would be correct if the code had compiled, and the subclass Alpha had been saved in its own file. In this case Java supplies an implicit call from the sub-class constructor to the no-args constructor of the super-class therefore line 12 causes Base to be output. Line 13 also causes Base to be output.
Option A is wrong. It would be correct if either the main class or the subclass had not been instantiated.
Option C is wrong. The code compiles.
Option D is wrong. There is output.
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What will be the output of the program?
public class ArrayTest
{
public static void main(String[ ] args)
{
float f1[ ], f2[ ];
f1 = new float[10];
f2 = f1;
System.out.println("f2[0] = " + f2[0]);
}
}
Answer: A
Option A is correct. When you create an array (f1 = new float[10];) the elements are initialises to the default values for the primitive data type (float in this case - 0.0), so f1 will contain 10 elements each with a value of 0.0. f2 has been declared but has not been initialised, it has the ability to reference or point to an array but as yet does not point to any array. f2 = f1; copies the reference (pointer/memory address) of f1 into f2 so now f2 points at the array pointed to by f1.
This means that the values returned by f2 are the values returned by f1. Changes to f1 are also changes to f2 because both f1 and f2 point to the same array.
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