Skip to content
  • Categories
  • Recent
  • Tags
  • Popular
  • World
  • Users
  • Groups
Skins
  • Light
  • Cerulean
  • Cosmo
  • Flatly
  • Journal
  • Litera
  • Lumen
  • Lux
  • Materia
  • Minty
  • Morph
  • Pulse
  • Sandstone
  • Simplex
  • Sketchy
  • Spacelab
  • United
  • Yeti
  • Zephyr
  • Dark
  • Cyborg
  • Darkly
  • Quartz
  • Slate
  • Solar
  • Superhero
  • Vapor
  • Default (No Skin)
  • No Skin
Collapse
Code Project
CODE PROJECT For Those Who Code
  • Home
  • Articles
  • FAQ
Community
  1. Home
  2. The Lounge
  3. bit operation challenge

bit operation challenge

Scheduled Pinned Locked Moved The Lounge
16 Posts 12 Posters 0 Views 1 Watching
  • Oldest to Newest
  • Newest to Oldest
  • Most Votes
Reply
  • Reply as topic
Log in to reply
This topic has been deleted. Only users with topic management privileges can see it.
  • 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

    R Offline
    R Offline
    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"

    1 Reply Last reply
    0
    • 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

      R Offline
      R Offline
      Robert Surtees
      wrote on last edited by
      #7

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

      A P S 3 Replies Last reply
      0
      • 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

        G 1 Reply Last reply
        0
        • 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

          1 Reply Last reply
          0
          • R Robert Surtees

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

            A Offline
            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

            G P 2 Replies Last reply
            0
            • 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

              J Offline
              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

              1 Reply Last reply
              0
              • 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

                G Offline
                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) { }

                1 Reply Last reply
                0
                • 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

                  P Offline
                  P Offline
                  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
                  Linkify!|Fold With Us!

                  1 Reply Last reply
                  0
                  • R Robert Surtees

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

                    P Offline
                    P Offline
                    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!

                    1 Reply Last reply
                    0
                    • 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

                      G Offline
                      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![^]

                      1 Reply Last reply
                      0
                      • R Robert Surtees

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

                        S Offline
                        S Offline
                        Sebastian Schneider
                        wrote on last edited by
                        #16

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

                        1 Reply Last reply
                        0
                        Reply
                        • Reply as topic
                        Log in to reply
                        • Oldest to Newest
                        • Newest to Oldest
                        • Most Votes

                        • Login
                        • Don't have an account? Register
                        • Login or register to search.
                        • First post
                          Last post
                        0
                        • Categories
                        • Recent
                        • Tags
                        • Popular
                        • World
                        • Users
                        • Groups