Objects and Collections - Finding the output Questions and Answers

31.

What will be the output of the program?

import java.util.*;

public class Trial31
{
    public static void main(String[] args)
    {
        Set<String> tags = new TreeSet<String>();
        tags.add("beta");
        tags.add("alpha");
        System.out.print(tags.first() + " ");
        tags.add(null);
        System.out.println(tags);
    }
}

A.alpha [alpha, beta]
B.alpha [null, alpha, beta]
C.alpha, then a NullPointerException
D.beta, then a NullPointerException

Answer: C

first returns the lowest element under the natural ordering, which is alpha, and that prints. Adding null then fails: a TreeSet must compare each new element against those present, and comparing null throws NullPointerException. A HashSet accepts a single null happily, needing only a hash and equality.

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32.

What will be the output of the program?

import java.util.*;

public class Case32
{
    public static void main(String[] args)
    {
        Map<String, String> m = new Hashtable<String, String>();
        m.put("k", null);
        System.out.println(m.size());
    }
}

A.The program prints 0
B.Compilation fails.
C.The program prints 1
D.A NullPointerException is thrown.

Answer: D

Hashtable is the legacy synchronised map and permits neither a null key nor a null value, throwing NullPointerException on either. HashMap allows one null key and any number of null values, which is the distinction being drawn.

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33.

What will be the output of the program?

import java.util.*;

public class Sketch33
{
    public static void main(String[] args)
    {
        Queue<Integer> pq = new PriorityQueue<Integer>(
            Arrays.asList(30, 10, 20));
        System.out.print(pq.poll());
        System.out.print(pq.poll());
        System.out.print(pq.poll());
    }
}

A.The program prints 102030
B.The program prints 203010
C.The order is unspecified.
D.The program prints 302010

Answer: A

A PriorityQueue hands elements out in priority order, which for Integer is ascending, so successive polls give 10, 20 then 30 regardless of insertion order. Note that only removal is ordered: printing the queue itself would show internal heap order, which is not sorted.

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34.

What will be the output of the program?

import java.util.*;

public class Item34
{
    public static void main(String[] args)
    {
        Deque<String> stack = new ArrayDeque<String>();
        stack.push("a");
        stack.push("b");
        System.out.println(stack.pop() + stack.peek());
    }
}

A.The program prints ab
B.The program prints ba
C.The program prints aa
D.The program prints bb

Answer: B

push and pop both work at the head of a Deque, so this behaves as a stack: b goes on top and comes off first. peek then reports the new head without removing it, which is a. The result is therefore ba, last-in-first-out.

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35.

What will be the output of the program?

import java.util.*;

public class Sample35
{
    public static void main(String[] args)
    {
        List<Integer> l = new ArrayList<Integer>(
            Arrays.asList(1, 2, 3, 4));
        boolean changed = l.retainAll(Arrays.asList(2, 4, 6));
        System.out.println(changed + " " + l);
    }
}

A.The program prints true [1, 3]
B.false [1
C.The program prints true [2, 4]
D.The program prints true [2

Answer: C

retainAll keeps only the elements also present in the argument, so 1 and 3 go and 2 and 4 stay. The 6 is never added, because retainAll only ever removes. It returns true because the list changed, which is how the bulk operations report their effect.

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36.

What will be the output of the program?

import java.util.*;

public class Probe36
{
    public static void main(String[] args)
    {
        List<String> l = new ArrayList<String>(
            Arrays.asList("a", "b"));
        System.out.println(
            l.addAll(new ArrayList<String>()) + " " + l.size());
    }
}

A.The program prints true 0
B.The program prints false 0
C.The program prints true 2
D.The program prints false 2

Answer: D

addAll returns whether the collection changed as a result of the call. Adding an empty collection changes nothing, so it returns false and the size stays at 2. The return value reports a change, not success, which is the same convention add and remove follow.

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37.

What will be the output of the program?

import java.util.*;

public class Demo37
{
    public static void main(String[] args)
    {
        List<String> l = new ArrayList<String>(
            Arrays.asList("a", "b"));
        Object[] plain = l.toArray();
        String[] typed = l.toArray(new String[0]);
        System.out.println(plain.length + " " + typed[1]);
    }
}

A.The program prints 2 b
B.The program prints 0 b
C.Compilation fails.
D.The program prints 2 a

Answer: A

The no-argument toArray can only promise Object[], while the array-taking form uses the supplied type and returns String[], allocating a bigger array when the given one is too small. Both hold the same two elements, so the length is 2 and the second element is b.

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38.

What will be the output of the program?

import java.util.*;

public class Snippet38
{
    public static void main(String[] args)
    {
        List<String> l = Collections.emptyList();
        System.out.print(l.size());
        l.add("a");
    }
}

A.The program prints 1
B.0, then an UnsupportedOperationException
C.The program prints 0
D.A NullPointerException is thrown.

Answer: B

emptyList returns an immutable shared instance, so the size prints as 0 and the add then throws UnsupportedOperationException. It exists to avoid allocating a new empty list, which is only safe because nothing can modify it.

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39.

What will be the output of the program?

import java.util.*;

public class Listing39
{
    public static void main(String[] args)
    {
        int[] a = {3, 1, 2};
        int[] b = {3, 1, 2};
        System.out.println(
            (a == b) + " " + Arrays.equals(a, b) + " " + a.equals(b));
    }
}

A.The program prints true true false
B.The program prints true true true
C.The program prints false true false
D.false false false

Answer: C

Arrays do not override equals, so both == and the inherited equals compare references and are false for two separately created arrays. Arrays.equals is the static helper that compares lengths and then elements, and it is true. The middle result is the only one that looks at contents.

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40.

What will be the output of the program?

import java.util.*;

public class Fragment40
{
    public static void main(String[] args)
    {
        int[] a = {5, 1, 4};
        int[] copy = Arrays.copyOfRange(a, 1, 3);
        Arrays.sort(a);
        System.out.println(Arrays.toString(copy) + " " + Arrays.toString(a));
    }
}

A.The program prints [1, 4] [5, 1, 4]
B.The program prints [5, 1] [1, 4, 5]
C.[1, 4, 5] [1, 4, 5]
D.The program prints [1, 4] [1, 4, 5]

Answer: D

copyOfRange takes the start index inclusive and the end exclusive, so it copies positions 1 and 2, giving [1, 4]. It is a copy rather than a view, so sorting the original afterwards leaves it untouched while a becomes [1, 4, 5]. Arrays.sort works in place and returns nothing.

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