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

Question

Lets 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?

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Related:

• How do you set, clear and toggle a single bit in C? (syntax varies, but the operations are the same)

Solution 1

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.

Solution 2

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

Solution 3

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

Comments

[Mithrax]

Right, but what would my mask be? 0x0000? Do you see what I mean? When doing AND I need to have a mask, how do I specify zero's with just those positions as zero?

[JP Alioto]

If you want to keep what you have there do AND with 1, if you want to erase what you have there, AND with 0.

[paxdiablo]

@staticsan, I expanded with the bit operations to make it clearer, then +1'ed you since the answer is now so much better :-) Hope you don't mind.

[paxdiablo]

@Varkhan, I think you mean ~ is the ones-complement. Twos-complement of n is -n (~n + 1 from memory).

[paxdiablo]

That would work with 0x0448 but not any other number - you need 0xfff1 to leave all the other bits untouched.

[chaos]

Yeah. Too bad there's no 'n' literal modifier.