Exercise: Java Lang Class

Questions for: Java Lang Class

What will be the output of the program? 
String x = "xyz";
x.toUpperCase(); /* Line 2 */
String y = x.replace('Y', 'y');
y = y + "abc";
System.out.println(y);
A:
abcXyZ
B:
abcxyz
C:
xyzabc
D:
XyZabc
Answer: C

Line 2 creates a new String object with the value "XYZ", but this new object is immediately lost because there is no reference to it. Line 3 creates a new String object referenced by y. This new String object has the value "xyz" because there was no "Y" in the String object referred to by x. Line 4 creates a new String object, appends "abc" to the value "xyz", and refers y to the result.

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What will be the output of the program? 
public class Test178 
{ 
    public static void main(String[] args) 
    {
        String s = "foo"; 
        Object o = (Object)s; 
        if (s.equals(o)) 
        { 
            System.out.print("AAA"); 
        } 
        else 
        {
            System.out.print("BBB"); 
        } 
        if (o.equals(s)) 
        {
            System.out.print("CCC"); 
        } 
        else 
        {
            System.out.print("DDD"); 
        } 
    } 
}
A:
AAACCC
B:
AAADDD
C:
BBBCCC
D:
BBBDDD
Answer: A
No answer description is available. Let's discuss.
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What will be the output of the program? 
public class Example 
{
    public static void main(String [] args) 
    {
        double values[] = {-2.3, -1.0, 0.25, 4};
        int cnt = 0;
        for (int x=0; x < values.length; x++) 
        {
            if (Math.round(values[x] + .5) == Math.ceil(values[x])) 
            {
                ++cnt;
            }
        }
        System.out.println("same results " + cnt + " time(s)");
    }
}
A:
same results 0 time(s)
B:
same results 2 time(s)
C:
same results 4 time(s)
D:
Compilation fails.
Answer: B

Math.round() adds .5 to the argument then performs a floor(). Since the code adds an additional .5 before round() is called, it's as if we are adding 1 then doing a floor(). The values that start out as integer values will in effect be incremented by 1 on the round() side but not on the ceil() side, and the noninteger values will end up equal.

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What will be the output of the program? 
public class ObjComp 
{
    public static void main(String [] args ) 
    {
        int result = 0;
        ObjComp oc = new ObjComp();
        Object o = oc;

        if (o == oc)  
            result = 1;
        if (o != oc)  
            result = result + 10;
        if (o.equals(oc) )  
            result = result + 100;
        if (oc.equals(o) )  
            result = result + 1000;

        System.out.println("result = " + result);
    }
}
A:
1
B:
10
C:
101
D:
1101
Answer: D

Even though o and oc are reference variables of different types, they are both referring to the same object. This means that == will resolve to true and that the default equals() method will also resolve to true.

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What will be the output of the program? 
public class BoolTest 
{
    public static void main(String [] args) 
    {
        int result = 0;

        Boolean b1 = new Boolean("TRUE");
        Boolean b2 = new Boolean("true");
        Boolean b3 = new Boolean("tRuE");
        Boolean b4 = new Boolean("false");

        if (b1 == b2)  /* Line 10 */
            result = 1;
        if (b1.equals(b2) ) /* Line 12 */
            result = result + 10;
        if (b2 == b4)  /* Line 14 */
            result = result + 100;
        if (b2.equals(b4) ) /* Line 16 */
            result = result + 1000;
        if (b2.equals(b3) ) /* Line 18 */
            result = result + 10000;

        System.out.println("result = " + result);
    }
}
A:
0
B:
1
C:
10
D:
10010
Answer: D

Line 10 fails because b1 and b2 are two different objects. Lines 12 and 18 succeed because the Boolean String constructors are case insensitive. Lines 14 and 16 fail because true is not equal to false.

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