Confusion with bitwise operations | and <<

Question

unsigned long long n = 0;
for (int i = 0; i <= 64; i+=2)
    n |= 1ULL << i;       //WHAT DOES THIS DO? AH!

I'm trying to wrap my head around what the third line of this code actually does. Someone please help clear this up!

Though does it really matter if all he's using is the bit pattern? — tskuzzy, Jun 14 '12 at 17:17
It actually invokes undefined behaviour to shift a 64 bit value by 64 bits -- only 0 to 63 are allowed. Better change the `i <= 64` to `i < 64`. — Secure, Jun 14 '12 at 19:33

score 6 · Accepted Answer · edited May 23 '17 at 10:28

That line sets the i^th bit of n.

1ULL is the integer 1 with type unsigned long long.
<< is a bitshift operator. 1ULL << i is equal to 2ⁱ, or in binary: 100...0 with i zeros.
n |= x; is a compound assignment operator. It is similar to writing n = n | x;.
The operator | is the bitwise OR operator.

Wikipedia has an example showing how bitwise OR operator works in the general case:

   0101 (decimal 5)
OR 0011 (decimal 3)
 = 0111 (decimal 7)

Related

score 1 · Answer 2 · answered Jun 14 '12 at 17:09

1

The |= 1ULL << i simply means set the ith bit. The for loops over every second bit so every other bit in the 64 bit unsigned long long will be set to 1.

In other words, you will get a bit pattern like ...0101010101.

answered Jun 14 '12 at 17:09

tskuzzy

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score 1 · Answer 3 · answered Jun 14 '12 at 17:09

1

It's left shifting 1 by i places, and OR'ing the result with n. Effectively it's setting bit i in n.

answered Jun 14 '12 at 17:09

peacemaker

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score 1 · Answer 4 · edited May 10 '16 at 12:16

1

it binary ORs n with the value 1 bit shifted by the value of i.

I believe the (binary) value of n would be:

0101010101010101010101010101010101010101010101010101010101010101

when the loop completed, although I haven't tested it..

edited May 10 '16 at 12:16

Antonio

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answered Jun 14 '12 at 17:10

pdriegen

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I didn't see the `i+=2` so I deleted my answer. – bluevector Jun 14 '12 at 17:11

Donnie · Answer 5 · 2012-06-14T17:19:07.530

1

n |= 1ULL << i;       //WHAT DOES THIS DO? AH!

On the right hand side you have "1ULL" which is a constant 1 unsigned long long. You are left bit shifting "1ULL" i number of times. The result of the left shifting of "1ULL" will then be matched with n to perform a bitwise OR. So n will be set to (n | (1ULL << i)).

This entire line or operation is setting the ith bit of n to a 1.

edited Jun 14 '12 at 17:19

answered Jun 14 '12 at 17:13

Donnie

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Confusion with bitwise operations | and <<

5 Answers5