Thread: Bitwise Questions

(var1 & 0) is always zero, so it won't do what you ask for.

Code:

!var1

will give the result you want. Of course, this is not a bitwise operator, but it's most likely possible to solve (in the compiled code) without branches, which I presume is your actual goal - whethter you use if-statements or not shouldn't be a goal in optimization, but avoiding branches/jumps certainly SHOULD be.

The second one can be solved with a simple and-statement - I'll let you post your attempt first - because that way, you will LEARN SOMETHING more than "I can post something and get an answer that I can paste into my code" - I think you know how to do the latter already.

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Mats

Originally Posted by someprogr

i = var1 | 128 ?

Nope. Use AND, not OR.

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Mats

Originally Posted by matsp

Nope. Use AND, not OR.

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Mats

actually, wouldnt shifting it to the right and isolating it be faster? i.e. i = var1 >> 7

Originally Posted by someprogr

actually, wouldnt shifting it to the right and isolating it be faster?

Not if you want to have i = 128 when var1 has the bit no 7 set. If you want i = 1 when bit 7 is set, then shifting it would be the right thing to do. Shifting is never faster than and, but it is slower on at least older/smaller processors - modern high-end processors, and the result is probably the same amount of time to shift or mask the bit.

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Mats

Originally Posted by itCbitC

bitwise AND'ing with octal 0100 ought to do the job as in

Code:

i = (var1 & 0100) ? 128 : 0;

Ehm, yes, if bit 7 is the bit worth 64, then I would agree that this would achieve the desired effect - but the terniary operator is often implmented as a if-statement, so may lead to a branch, so if this is REALLY what you want to do, then:

Code:

i = (var1 & 64) << 1;

will achieve the same thing, guaranteed to be without a branch.

Using octal will achieve nothing special here.

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Mats