Java.lang Class - Finding the output Questions and Answers

21.

What will be the output of the program?

public class Trial21
{
    Trial21()
    {
        Integer a = 127, b = 127;
        Integer c = 128, d = 128;
        System.out.println((a == b) + " " + (c == d));
    }

    public static void main(String[] args)
    {
        new Trial21();
    }
}

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

Answer: A

Autoboxing an int between -128 and 127 must return a cached Integer, so a and b reference the same object and the first test is true. 128 is outside that guaranteed range, so each boxing may create a new object and the second is false. equals would be true in both cases.

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22.

What will be the output of the program?

public class Case22
{
    public static void main(String[] args) throws Exception
    {
        Integer x = 1000;
        int y = 1000;
        System.out.println(x == y);
    }
}

A.The program prints false
B.The program prints true
C.Compilation fails.
D.A NullPointerException is thrown.

Answer: B

Comparing a wrapper with a primitive unboxes the wrapper first, so this is an int comparison of 1000 with 1000 and the answer is true. The cache is irrelevant because no reference comparison happens. Only wrapper against wrapper compares identity.

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23.

What will be the output of the program?

public class Sketch23
{
    private static void report()
    {
        int parsed = Integer.parseInt("0011");
        Integer boxed = Integer.valueOf("11");
        System.out.println(parsed + " " + boxed);
    }

    public static void main(String[] args)
    {
        report();
    }
}

A.The program prints 11 3
B.The program prints 0011 11
C.The program prints 11 11
D.The program prints 3 11

Answer: C

parseInt reads base ten unless another radix is given, so the leading zeros are simply insignificant digits and the value is 11, not the octal or binary reading of 0011. valueOf does the same and boxes the result. Passing 2 as a radix is what would give 3.

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24.

What will be the output of the program?

public class Item24
{
    void execute()
    {
        String bits = Integer.toBinaryString(10);
        String hex = Integer.toHexString(255);
        System.out.println(bits + " " + hex);
    }

    public static void main(String[] args)
    {
        Item24 demo = new Item24();
        demo.execute();
    }
}

A.1010 and FF, in upper case
B.0b1010 and 0xff, with prefixes
C.10 and 255, unchanged
D.1010 and ff, in lower case

Answer: D

toBinaryString and toHexString return the digits alone, with no 0b or 0x prefix, and the hexadecimal letters are lower case. So 10 becomes 1010 and 255 becomes ff. Upper case would need toUpperCase() on the result.

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25.

What will be the output of the program?

public class Sample25
{
    public static void main(String[] args)
    {
        int biggest = Integer.MAX_VALUE;
        int wrapped = biggest + 1;
        System.out.println(wrapped);
    }
}

A.The program prints -2147483648
B.The program prints 2147483647
C.An ArithmeticException is thrown.
D.The program prints 2147483648

Answer: A

int arithmetic wraps silently on overflow rather than throwing or widening, so one past the largest int is the smallest one. The result is -2147483648. Adding 1L instead would promote the expression to long and give the mathematical answer.

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26.

What will be the output of the program?

public class Probe26
{
    static void show()
    {
        Integer i = null;
        System.out.println(i == null);
        int j = i;
    }

    public static void main(String[] args)
    {
        show();
    }
}

A.Compilation fails.
B.A NullPointerException is thrown after printing true
C.A NullPointerException is thrown before anything is printed
D.The program prints true and ends normally

Answer: B

Comparing a wrapper reference against the null literal does not unbox, so the print succeeds and reports true. The assignment on the next line does unbox, calling intValue on null, and throws NullPointerException. Output already written is not undone by the later failure.

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27.

What will be the output of the program?

public class Demo27
{
    void run()
    {
        boolean shouty = Boolean.parseBoolean("TRUE");
        boolean affirmative = Boolean.parseBoolean("yes");
        System.out.println(shouty + " " + affirmative);
    }

    public static void main(String[] args)
    {
        new Demo27().run();
    }
}

A.An IllegalArgumentException is thrown.
B.The program prints true true
C.The program prints true false
D.The program prints false false

Answer: C

parseBoolean compares against the text true ignoring case, so TRUE gives true. Anything else, including yes, gives false rather than throwing. The method has no notion of other affirmative words and never reports a parse failure.

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28.

What will be the output of the program?

public class Snippet28
{
    public static void main(String... args)
    {
        double d = 0.0 / 0.0;
        System.out.println(d + " " + (d == d));
    }
}

A.The program prints 0.0 true
B.An ArithmeticException is thrown.
C.NaN true
D.NaN false

Answer: D

Zero divided by zero in floating point is NaN rather than an error, and NaN is specified to compare unequal to every value including itself. So the value prints as NaN and the identity test is false. Integer division by zero would have thrown instead.

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29.

What will be the output of the program?

public class Listing29
{
    Listing29()
    {
        double up = 1.0 / 0;
        double down = -1.0 / 0;
        System.out.println(up + " " + down);
    }

    public static void main(String[] args)
    {
        new Listing29();
    }
}

A.Infinity -Infinity
B.NaN NaN
C.An ArithmeticException is thrown.
D.The program prints 0.0 -0.0

Answer: A

Dividing a non-zero double by zero gives a signed infinity rather than throwing, because the floating-point types have values for it. Only zero over zero is indeterminate and gives NaN. The same expressions with int operands would throw ArithmeticException.

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30.

What will be the output of the program?

public class Fragment30
{
    public static void main(String[] args) throws Exception
    {
        System.out.println(0.1 + 0.2 == 0.3);
        System.out.println(0.1 + 0.2);
    }
}

A.The program prints true, then 0.3
B.false, then 0.30000000000000004
C.The program prints false, then 0.3
D.true, then 0.30000000000000004

Answer: B

Neither operand is exactly representable in binary, and their sum lands slightly above three tenths, so the equality is false and printing the sum shows the excess digits. Floating-point comparisons need a tolerance rather than exact equality.

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Java.lang Class - Finding the output Questions and Answers

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