Java Programming - Operators and Assignments - Discussion

Discussion Forum : Operators and Assignments - Finding the output (Q.No. 14)
14.
What will be the output of the program?
class BoolArray 
{
    boolean [] b = new boolean[3];
    int count = 0;

    void set(boolean [] x, int i) 
    {
        x[i] = true;
        ++count;
    }

    public static void main(String [] args) 
    {
        BoolArray ba = new BoolArray();
        ba.set(ba.b, 0);
        ba.set(ba.b, 2);
        ba.test();
    }

    void test() 
    {
        if ( b[0] && b[1] | b[2] )
            count++;
        if ( b[1] && b[(++count - 2)] )
            count += 7;
        System.out.println("count = " + count);
    }
}
count = 0
count = 2
count = 3
count = 4
Answer: Option
Explanation:

The reference variables b and x both refer to the same boolean array. count is incremented for each call to the set() method, and once again when the first if test is true. Because of the && short circuit operator, count is not incremented during the second if test.

Discussion:
31 comments Page 1 of 4.

Harshit Rochiramani said:   7 years ago
When the second call is made i.e. if ( b[1] && b[(++count - 2)] ) => if ( FALSE && b[(++count - 2)]).

Another part is never executed because of compiler optimisation in this case JVM hence it outputs 3!

if you write if (b[(++count - 2)] && b[1]).
It will output 4.
(2)

Sanjai Krishnan said:   7 years ago
In the main method, we pass values for the index for 0 and 2.

Hence X[0]=true ,[2]= true. Default X[1]= false... Hence by changing the X[0], X[2] to true the count will become as count=2.

Next when the test method is called from the main method. First, if the condition is like (T&&F|T) which is false and hence count value is not incremented and remains as Count=2.

Second if condition is like (T && b[(++ count-2)])... Ie b[3-2] = b[1 ].
Since it is ++Count ... count = 3.... Where b[1] is false.

It looks like (T&&F ) which is False and hence count+=7 is not executed.
Thus the count=3 is to be displayed.
(1)

Sanjai Krishnan said:   7 years ago
In main method , we pass values for index for 0 and 2.

Hence X[0]=true ,[2]= true. Default X[1]= false... Hence by changing the X[0], X[2] to true the count will become as count=2.

Next when test method is called from main method.

First ,if condition is like (T&&F|T) which is false and hence count value is not incremented and remains as Count=2.

Second,if condition is like (T && b[(++ count-2)])... Ie b[3-2] = b[1 ].
Since it is ++Count ... count = 3.... Where b[1] is false.
It looks like (T&&F ) which is False and hence count+=7.
Thus the count=3 is to be displayed.

Anant said:   7 years ago
I am not getting it. Can anyone help me?

Preethi said:   7 years ago
Thank you for the clear explanation @Jndsln.

Parth Soni said:   8 years ago
Can anyone explain how b and x refer to the same array because x's scope is in set method only and it will not update the array b right ?

Sainath said:   8 years ago
You are absolutely right Thanks, @Jndlsnl.

Bittu said:   8 years ago
Yes, @Jndlsnl.

You are right. Thanks a lot for writing down if values.

Swagatika said:   9 years ago
Please anyone explain the compleat program.

Ravi said:   9 years ago
Thanks @Huy.


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