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bit operation challenge

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  • A 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.

    I can imagine the sinking feeling one would have after ordering my book, only to find a laughably ridiculous theory with demented logic once the book arrives - Mark McCutcheon

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    Chris Maunder
    wrote on last edited by
    #4

    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

    CodeProject.com : C++ MVP

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    • A alex barylski

      It's technical in nature but doesn't allpy to any specific language...so I'm asking it here: 1) I have a variable which is three bits long. 2) I need a quick bitwise operation (single statement) that returns FALSE if anymore than one bit is set (no ternary operator). Mutex type check I guess :) The user should only ever specify one field, but they might do otherwise which would yield inaccurate results. So say: 011 is passed, I need a quick check to confirm it's more than a single bit set and thus returning false, so the function can be unit tested I can't think of any combination of bitwise operators which would yield such a result - is it possible? Cheers :)

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

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      Lost User
      wrote on last edited by
      #5

      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.

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      • A alex barylski

        It's technical in nature but doesn't allpy to any specific language...so I'm asking it here: 1) I have a variable which is three bits long. 2) I need a quick bitwise operation (single statement) that returns FALSE if anymore than one bit is set (no ternary operator). Mutex type check I guess :) The user should only ever specify one field, but they might do otherwise which would yield inaccurate results. So say: 011 is passed, I need a quick check to confirm it's more than a single bit set and thus returning false, so the function can be unit tested I can't think of any combination of bitwise operators which would yield such a result - is it possible? Cheers :)

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

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        Roger Wright
        wrote on last edited by
        #6

        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"

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        • A alex barylski

          It's technical in nature but doesn't allpy to any specific language...so I'm asking it here: 1) I have a variable which is three bits long. 2) I need a quick bitwise operation (single statement) that returns FALSE if anymore than one bit is set (no ternary operator). Mutex type check I guess :) The user should only ever specify one field, but they might do otherwise which would yield inaccurate results. So say: 011 is passed, I need a quick check to confirm it's more than a single bit set and thus returning false, so the function can be unit tested I can't think of any combination of bitwise operators which would yield such a result - is it possible? Cheers :)

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

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          Robert Surtees
          wrote on last edited by
          #7

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

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          • A 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.

            I can imagine the sinking feeling one would have after ordering my book, only to find a laughably ridiculous theory with demented logic once the book arrives - Mark McCutcheon

            A Offline
            A Offline
            alex barylski
            wrote on last edited by
            #8

            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

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            • C 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

              CodeProject.com : C++ MVP

              A Offline
              A Offline
              alex barylski
              wrote on last edited by
              #9

              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

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              • R Robert Surtees

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

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                A Offline
                alex barylski
                wrote on last edited by
                #10

                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

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                • A alex barylski

                  It's technical in nature but doesn't allpy to any specific language...so I'm asking it here: 1) I have a variable which is three bits long. 2) I need a quick bitwise operation (single statement) that returns FALSE if anymore than one bit is set (no ternary operator). Mutex type check I guess :) The user should only ever specify one field, but they might do otherwise which would yield inaccurate results. So say: 011 is passed, I need a quick check to confirm it's more than a single bit set and thus returning false, so the function can be unit tested I can't think of any combination of bitwise operators which would yield such a result - is it possible? Cheers :)

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

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                  J Offline
                  John R Shaw
                  wrote on last edited by
                  #11

                  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

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                  • A 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

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                    G Offline
                    gmarian
                    wrote on last edited by
                    #12

                    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) { }

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                    • A 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

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                      peterchen
                      wrote on last edited by
                      #13

                      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.

                      Developers, Developers, Developers, Developers, Developers, Developers, Velopers, Develprs, Developers!
                      We are a big screwed up dysfunctional psychotic happy family - some more screwed up, others more happy, but everybody's psychotic joint venture definition of CP
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                      • R Robert Surtees

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

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                        peterchen
                        wrote on last edited by
                        #14

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

                        Developers, Developers, Developers, Developers, Developers, Developers, Velopers, Develprs, Developers!
                        We are a big screwed up dysfunctional psychotic happy family - some more screwed up, others more happy, but everybody's psychotic joint venture definition of CP
                        Linkify!|Fold With Us!

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                        • A 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

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                          G Offline
                          Gary R Wheeler
                          wrote on last edited by
                          #15

                          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;

                          Fold With Us![^]

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                          • R Robert Surtees

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

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                            Sebastian Schneider
                            wrote on last edited by
                            #16

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

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