OOPS
August 29, 2024OOPS
August 29, 2024OOPS
| Question 79 |
What does the following java function perform ? (Assume int occupies four bytes of storage)
Public static int f(int a) { //Pre-conditions : a>0 and no overflow/underflow occurs int b=0; for(int i=0; i<32; i++) { b=b <<1; b= b | (a & 1); a = a >>> 1; // This is a logical shift } Return b; }
| Return the int that represents the number of 0’s in the binary representation of integer a. | |
| Return the int that represents the number of 1’s in the binary representation of integer a. | |
| Return the int that has the reversed binary representation of integer a.
| |
| Return the int that has the binary representation of integer a.
|
Question 79 Explanation:
→ Consider the a value is 5.
The initial value of b is 0 and b<<1 means b will get value of 2
b=b|(a&1) which means 2|(5&1) then b will get value “2”
a=a<<1 means value a will reduce to 3
→ Repeat the above procedure for 31 iterations.
→ Iteration-2: b=b<<1 then b is 4 and b=b|(a&1)= 4|(3&1) then b is 5 and a becomes 1
→ Iteration-3: b=b<<1 then b is 10 and b=b|(1&1) = 4|(3&1) then b is 11 and a becomes 0
→ And this procedure will repeat until “i” value is 32.
→ The final value is 11 (Number of 1’s in the result is 2) and the number of 1’s in the input 5 are 2.
The initial value of b is 0 and b<<1 means b will get value of 2
b=b|(a&1) which means 2|(5&1) then b will get value “2”
a=a<<1 means value a will reduce to 3
→ Repeat the above procedure for 31 iterations.
→ Iteration-2: b=b<<1 then b is 4 and b=b|(a&1)= 4|(3&1) then b is 5 and a becomes 1
→ Iteration-3: b=b<<1 then b is 10 and b=b|(1&1) = 4|(3&1) then b is 11 and a becomes 0
→ And this procedure will repeat until “i” value is 32.
→ The final value is 11 (Number of 1’s in the result is 2) and the number of 1’s in the input 5 are 2.
Correct Answer: C
Question 79 Explanation:
→ Consider the a value is 5.
The initial value of b is 0 and b<<1 means b will get value of 2
b=b|(a&1) which means 2|(5&1) then b will get value “2”
a=a<<1 means value a will reduce to 3
→ Repeat the above procedure for 31 iterations.
→ Iteration-2: b=b<<1 then b is 4 and b=b|(a&1)= 4|(3&1) then b is 5 and a becomes 1
→ Iteration-3: b=b<<1 then b is 10 and b=b|(1&1) = 4|(3&1) then b is 11 and a becomes 0
→ And this procedure will repeat until “i” value is 32.
→ The final value is 11 (Number of 1’s in the result is 2) and the number of 1’s in the input 5 are 2.
The initial value of b is 0 and b<<1 means b will get value of 2
b=b|(a&1) which means 2|(5&1) then b will get value “2”
a=a<<1 means value a will reduce to 3
→ Repeat the above procedure for 31 iterations.
→ Iteration-2: b=b<<1 then b is 4 and b=b|(a&1)= 4|(3&1) then b is 5 and a becomes 1
→ Iteration-3: b=b<<1 then b is 10 and b=b|(1&1) = 4|(3&1) then b is 11 and a becomes 0
→ And this procedure will repeat until “i” value is 32.
→ The final value is 11 (Number of 1’s in the result is 2) and the number of 1’s in the input 5 are 2.