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.

Biswa said:   1 decade ago
Initially count is zero. The set method is call two times, so it 2. How we get 3 please explain.

Samir said:   1 decade ago
When test method call at that time count is increment at one more time.

Gopi said:   1 decade ago
But why it is incremented only in first if in test method ?

Jndlsnl@gmail.com said:   1 decade ago
Bcz in if ( b[0] && b[1] | b[2] ) ---> if(true && false | true) ----> if(true) so count++ =3.

but if ( b[1] && b[(++count - 2)] ) ----> if(false && b[(3-2)])---> if(false && false) ----> if(false) ----> so it will not execute.

Correct me if I am wrong.

Thanvir513@yahoo.com said:   1 decade ago
Why b[0]=true, b[1]=false and b[2]=true ? please explain clearly.

Shipra said:   1 decade ago
ba.set(ba.b, 0); will set b[0]=true and ba.set(ba.b, 2); will set b[2]=true.. n b[a]=0 so we take it as false.
now in test method first if statement will return true because (true & false) || true = true;
now in second if statement first statement b[1] is false so next statement will not be executed ..because there is AND operator does not check second condition if first is false..so count value will not be incremented n will remain 3.

Gourav Soni said:   1 decade ago
@[jndlsnl@gmail.com] :
in second if condition we have used ++count...which should increase count by 1 ...so count shud be 4 ..ryt??

Varsha said:   1 decade ago
@gourav : evn i had the same doubt but after reading shipras explanation i am convinced and my doubt is resolved.....

Ashi said:   1 decade ago
b[1]=false..how??

Nitish said:   1 decade ago
Ashi:-because by default the value for boolean will be false.
Gourav Soni and Varsha:-its working like:-
if ( b[1] && b[(++count - 2)] )
count += 7;
if(false && ----- )
here b[1] is false.So.. if the value for left operand is false then it won't check for right operand and will go to next line.that's y count won't be incremented further.
i hope it will be helpful for you.


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