Exercise: Operators And Assignments

Questions for: Operators And Assignments

What will be the output of the program? 
class SC2 
{
    public static void main(String [] args) 
    {
        SC2 s = new SC2();
        s.start();
    }

    void start() 
    {
        int a = 3;
        int b = 4;
        System.out.print(" " + 7 + 2 + " ");
        System.out.print(a + b);
        System.out.print(" " + a + b + " ");
        System.out.print(foo() + a + b + " ");
        System.out.println(a + b + foo());
    }

    String foo() 
    {
        return "foo";
    }
}
A:
9 7 7 foo 7 7foo
B:
72 34 34 foo34 34foo
C:
9 7 7 foo34 34foo
D:
72 7 34 foo34 7foo
Answer: D

Because all of these expressions use the + operator, there is no precedence to worry about and all of the expressions will be evaluated from left to right. If either operand being evaluated is a String, the + operator will concatenate the two operands; if both operands are numeric, the + operator will add the two operands.

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What will be the output of the program? 
class SSBool 
{
    public static void main(String [] args) 
    {
        boolean b1 = true;
        boolean b2 = false;
        boolean b3 = true;
        if ( b1 & b2 | b2 & b3 | b2 ) /* Line 8 */
            System.out.print("ok ");
        if ( b1 & b2 | b2 & b3 | b2 | b1 ) /*Line 10*/
            System.out.println("dokey");
    }
}
A:
ok
B:
dokey
C:
ok dokey
D:
No output is produced
Answer: B

The & operator has a higher precedence than the | operator so that on line 8 b1 and b2 are evaluated together as are b2 & b3. The final b1 in line 10 is what causes that if test to be true. Hence it prints "dokey".

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What will be the output of the program? 
class Bitwise 
{
    public static void main(String [] args) 
    {
        int x = 11 & 9;
        int y = x ^ 3;
        System.out.println( y | 12 );
    }
}
A:
0
B:
7
C:
8
D:
14
Answer: D

The & operator produces a 1 bit when both bits are 1. The result of the & operation is 9. The ^ operator produces a 1 bit when exactly one bit is 1; the result of this operation is 10. The | operator produces a 1 bit when at least one bit is 1; the result of this operation is 14.

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What will be the output of the program? 
class Test 
{
    public static void main(String [] args) 
    {
        int x= 0;
        int y= 0;
        for (int z = 0; z < 5; z++) 
        {
            if (( ++x > 2 ) || (++y > 2)) 
            {
                x++;
            }
        }
    System.out.println(x + " " + y);
    }
}
A:
5 3
B:
8 2
C:
8 3
D:
8 5
Answer: B

The first two iterations of the for loop both x and y are incremented. On the third iteration x is incremented, and for the first time becomes greater than 2. The short circuit or operator || keeps y from ever being incremented again and x is incremented twice on each of the last three iterations.

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What will be the output of the program? 
class Test 
{
    public static void main(String [] args) 
    {
        int x= 0;
        int y= 0;
        for (int z = 0; z < 5; z++) 
        {
            if (( ++x > 2 ) && (++y > 2)) 
            {
                x++;
            }
        }
        System.out.println(x + " " + y);
    }
}
A:
5 2
B:
5 3
C:
6 3
D:
6 4
Answer: C

In the first two iterations x is incremented once and y is not because of the short circuit && operator. In the third and forth iterations x and y are each incremented, and in the fifth iteration x is doubly incremented and y is incremented.

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