CODEWITHSUNDEEP
language-agnosticbit-manipulation
Mithrax
Mithrax

Reputation: 7809

How do I set or clear the first 3 bits using bitwise operations?

Let’s say I have a number like 0x448. In binary this is 0100 0100 1000.

How do I set the bits 1, 2 and 3 to either all 0's or all 1's using bit-wise operations? When I say the first three, I'm counting the rightmost bit as the zero bit.

So, for example

Bits as 1's:

b12            b0
  0100 0100 1110
            ^^^

Bits as 0's:

b12            b0
  0100 0100 0000
            ^^^

I'm guessing that to set them to 1's I use bit-wise OR with a mask of 14 (0x000e)? But if that is the case, how do I do something similar for clearing the bits?

Related:

Upvotes: 10

Views: 15389

Answers (7)

Vlad
Vlad

Reputation: 4525

Let’s make sure that bits are counted from 0 starting at the right side or least significant bit and going left. Then observe this:

 1 <<p puts 1 at the position p, for example

 1 <<3     = 00001000

 1 <<p - 1 puts 1 at all positions up to p, exclusive
 1 <<3-1=00000111

The last step produced a mask to clear bits from the most significant to p inclusive. You can invert it with tilde to clear the other half.

Upvotes: 0

Chand
Chand

Reputation: 87

For clearing the bits, use AND with 0x440.

Upvotes: 0

Varkhan
Varkhan

Reputation: 16761

Supposing you have a mask m with bits set to 1 for all the bits you want to set or clear, and 0 otherwise:

  • clear bits: x & (~m)
  • set bits: x | m
  • flip bits: x ^ m

If you are only interested in one bit, in position p (starting at 0), the mask is simple to express m = 1 << p

Note that I am using C-style conventions, where:

  • ~ is the 1-complement: ~10001010 = 01110101
  • & is the bitwise AND
  • | is the bitwise OR
  • ^ is the bitwise XOR
  • << is the left bit shift: 10001010 << 2 = 00101000

Upvotes: 4

chaos
chaos

Reputation: 124287

number &= ~0xe

Upvotes: 0

staticsan
staticsan

Reputation: 30555

You have the bit setting correct: OR with the mask of the bits you want to set.

Bit clearing bits is very similar: AND with the ones-complement of the bits you want cleared.

Example: Word of 0x0448.

Settings bits 1, 2 and 3 would be Word OR 0x000e:

    0000 0100 0100 1000 = 0x0448
 OR 0000 0000 0000 1110 = 0x000e
    ---- ---- ---- ----
  = 0000 0100 0100 1110 = 0x044e

Clearing bits 1, 2 and 3 would be Word AND 0xfff1:

    0000 0100 0100 1000 = 0x0448
AND 1111 1111 1111 0001 = 0xfff1
    ---- ---- ---- ----
  = 0000 0100 0100 0000 = 0x0440

Elaborating on the ones-complement, the AND pattern for clearing is the logical NOT of the OR pattern for setting (each bit reveresed):

 OR 0000 0000 0000 1110 = 0x000e
AND 1111 1111 1111 0001 = 0xfff1

so you can use your favorite language NOT operation instead of having to figure out both values.

Upvotes: 19

CookieOfFortune
CookieOfFortune

Reputation: 13974

Assuming your OR 0x14 is correct, clearing would be:

AND (NOT 0x14)

Upvotes: 0

JP Alioto
JP Alioto

Reputation: 45117

OR with 1 is always true; AND with 0 is always false. :)

Upvotes: 1

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