bit operation challenge

Anton Afanasyev

well, if you have the value of all three bits, then the following would do: answer = ((not b) AND (not c)) OR ((not a) AND (not b) AND c) OR ((not a) AND b AND (not c)) where a, b, c are your three bits. I'm not sure if this can be simplified further, but it would do the trick.

:badger:

Andy Brummer

I'm not sure what you are asking for. (x & 111b) > 0 will tell you if one or more of the 3 bits are set. Using and is essentially a bit mask.

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Chris Maunder

that yields true if 1 or more bits are set. How do you check if 1 and only 1 bit is set? (apart from "if value=1 or value=2 or value=4" ;))

cheers, Chris Maunder

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Lost User

How's about this? !(((a^110b)+((a^101b)>>1)+((a^011b)>>2))>>1) Edit - maybe I should explain a bit! Essentially just sum the binary digits (what the xor's and shifts are doing), the most you can have is 3 (or 11b) since we're limited to 3 binary digits. Shift that right one place and anything 2 or more leaves a 1 in the right most bit which, when not'd leaves a 0 (false). Anything else leaves a 1 (true). Edit again: This can be simplified !(((a^110b)+((a^101b)>>1)+(a>>2))>>1) since the effect of shifting two spots right removes the need for the xor. Cheers, Drew.

Roger Wright

Drew came up with the only solution I like (though I played with several less elegant forms), but another approach I like is to define your flag variable as a bitset, then use the count operator when you need to test it.

"A Journey of a Thousand Rest Stops Begins with a Single Movement"

Robert Surtees

b & (b - 1) is false for all powers of two

alex barylski

I wanted to see if it was possible to check if *only* a single bit was set, nothing more, nothing less otherwise return FALSE. Not using anything but a single complicated bitwise statement (no ternary, conditionals, switch, etc).

I finally got a job doing something I enjoy and I"m good at...15 hour days seem like play time :P

alex barylski

Yes...yeah exactly :)

I finally got a job doing something I enjoy and I"m good at...15 hour days seem like play time :P

alex barylski

Hhuh... i'm missing something, you need to explain more...this does the same as above? :(

I finally got a job doing something I enjoy and I"m good at...15 hour days seem like play time :P

John R Shaw

Here are a couple of more for you: (((x&1) + ((x&2)>>1) + ((x&4)>>2)) == 1); ((x&7) && !((x&3) || (x&5) || (x&6));

Hockey wrote:

I finally got a job doing something I enjoy and I"m good at...15 hour days seem like play time

I still enjoy it when I work on something new (with no pressure). ;) Other times it is: (1) I know the problem, (2) I know the solution, (3) Blast! Now I have to type all that before tomorrow. :sigh:

INTP "Program testing can be used to show the presence of bugs, but never to show their absence."Edsger Dijkstra

gmarian

Now that's the elegant solution! Try it for some numbers: 010b: 010b - 1 = 001b 010b & 001b = 0 100b: 100b - 1 = 010b 100b & 010b = 0 110b: 110b - 1 = 101b 110b * 101b = 100b so, if (bitFlag & (bitFlag - 1) == 0) { }

peterchen

Distinguish three cases: power of two, zero, and all other values You understand binary notation and bitwise operators? If b is a power of two (b = 2^N): b-1 is a number with all bits 0..N-1 set (e.g. for 2^3=8=1000b, 8-1 = 0111b) In this casethere are not bits set in both b and (b-1), so the '&' yields 0 b=0: 0 '&' (anything) remains zero For the rest: they can be represented as b = 2^N+g with a g>0 && g<2^N since g>0, both b and 8b-1) will have the bit^N set, so the '&' operation will return a non-zero value.

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peterchen

You beat me :mad: so I voted you a five, so you stick out ;)

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Gary R Wheeler

inline bool SingleBitOf3(unsigned X)
{
return (X & 0x04) || (X & 0x02) || (X & 0x01);
}

Kinda icky, mixing bitwise operators and logical operators in the same expression, but hey.

Software Zen: delete this;

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Sebastian Schneider

Brilliant. I am gonna kidnap you and force you to programm stuff in my basement.