0

I have the following code

int n = 50;
while(n) {                         //1
    if(n & 1) cout << "1" << endl; //2 
    //right shift the number so n will become 0 eventually and the loop will terminate
    n >>= 1;                       //3
}

When we use bitwise and 1 (& 1) with a number we get back the same number. Now my question is how does c++ evaluates the following expression: n & 1. Since:

  n = 50
  In binary form 50 is:            110010
  If we bitwise 1 then we get:  AND     1 = 110010
  Now in c++ (2) the expression evaluates like this:
  Instead of getting the whole sequence of bits (110010) bitwise anded with 1    
  it evaluates only the number of right bits we bitwise. In my example:
  n=50, 110010, use n & 1 ==> 0 AND 1 instead of 110010 AND 1.

Is there a reason that c++ treats the bitwise and like this? My guess would be it has to do with the compiler ?

Improve this question
4
  • 2
    But remember that 1 is actually 00000001 in binary, and bitwise operations use all the bits. Commented May 11, 2015 at 10:23
  • 1
    The same BIT VALUE, single bit, not number (which in general is many bits), in "When we use bitwise and 1 (& 1) with a number we get back the same number" Commented May 11, 2015 at 10:24
  • oh now i see so when i use n & 1 i actually get this: 110010 AND 000001 ? Commented May 11, 2015 at 10:26
  • @KostasRim that's literally what it is saying, not some sort of "I wrote this but I actually get this other thing". Commented May 11, 2015 at 10:31

2 Answers 2

Reset to default
3

When we use bitwise and 1 (& 1) with a number we get back the same number.

No we don't. We get back the number consisting of the bits that are set in both the original number and in 1. Since only the lowest bit of 1 is set, the result is the lowest bit of the original number.

Now my question is how does c++ evaluates the following expression: n & 1.

If n is 50, then in binary:

n:    110010
1:    000001
n&1:  000000 // no bits set in both

If n is 51, then in binary:

n:    110011
1:    000001
n&1:  000001 // one bit set in both
Improve this answer
Sign up to request clarification or add additional context in comments.

1 Comment

Thank you very much for your answer, i will mark it as solved.
2

From Wikipedia:

The bitwise AND operator is a single ampersand: &. It is just a representation of AND which does its work on the bits of the operands rather than the truth value of the operands. Bitwise binary AND does the logical AND (as shown in the table above) of the bits in each position of a number in its binary form.

In your example 110010 & 1, 1 is considered as 000001, and then each bit is anded and you get the result. In fact, I use this method: 1&number to check for even and odd numbers. This is how:

if(1 & num)
  printf("it is odd");
else 
  printf("it is even");

This is how it works: suppose you have an 8 bit number. Now, the 8 bit notation of 1 will be 00000001.

If I now perform and on each bit, for all the first seven bits I will get 0, because it will be 0 & anything will be 0. Now, the last bit of 1 is 1. So, if my number also has last bit as 1, then 1 & 1 = 1, and if my last bit is 0, then 1 & 0 = 0.

When will the last bit in my number be 1? And when 0? When converting to decimal form, the last bit is multiplied by 20. And, 20 = 1. If this 1 is multiplied with 1, we get an odd number, and if it is multiplied with 0, we get an even number.

Improve this answer

Comments

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.