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  4. What does #pragma pack(0) do

What does #pragma pack(0) do

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  • Y yu jian

    The msdn says #pragma pack(n) will change current alignment value to n. But if n is zero, what will do?

    C Offline
    C Offline
    CPallini
    wrote on last edited by
    #11

    Why don't you continue reading MSDN? #pragma pack documentation[^] states:

    Valid values are 1, 2, 4, 8, and 16.

    Hence 0 is 'not valid' (I wouldn't try to make assumptions on a value marked as such).

    Veni, vidi, vici.

    Y 1 Reply Last reply
    0
    • Y yu jian

      The msdn says #pragma pack(n) will change current alignment value to n. But if n is zero, what will do?

      L Offline
      L Offline
      Lost User
      wrote on last edited by
      #12

      yu-jian wrote:

      But if n is zero, what will do?

      Add /WX to your compiler settings and recompile. You should pay more attention to compiler warnings. :) Best Wishes, -David Delaune

      Y 1 Reply Last reply
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      • L Lakamraju Raghuram

        #pragma pack(0)

        #include
        using namespace std;

        struct Test
        {
        char a;
        int i;
        };

        void main()
        {
        cout<

        I am using VS2008 SP1. The build is x86.
        Now guess the result of sizeof(Test) ??

        Y Offline
        Y Offline
        yu jian
        wrote on last edited by
        #13

        8 in vs2008 sp1

        1 Reply Last reply
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        • C CPallini

          Why don't you continue reading MSDN? #pragma pack documentation[^] states:

          Valid values are 1, 2, 4, 8, and 16.

          Hence 0 is 'not valid' (I wouldn't try to make assumptions on a value marked as such).

          Veni, vidi, vici.

          Y Offline
          Y Offline
          yu jian
          wrote on last edited by
          #14

          Just after read MSDN, I found that the case n=0 is ignored. I do not know why.

          enhzflepE 1 Reply Last reply
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          • Y yu jian

            Just after read MSDN, I found that the case n=0 is ignored. I do not know why.

            enhzflepE Offline
            enhzflepE Offline
            enhzflep
            wrote on last edited by
            #15

            Simply because it's not explicitly handled. Without giving the directive a special meaning for n=0, it makes perfect sense - it keeps the use of the directive consistent.

            1 Reply Last reply
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            • L Lost User

              yu-jian wrote:

              But if n is zero, what will do?

              Add /WX to your compiler settings and recompile. You should pay more attention to compiler warnings. :) Best Wishes, -David Delaune

              Y Offline
              Y Offline
              yu jian
              wrote on last edited by
              #16

              There is a error that Visual Studio 2008 only supports 1, 2, 4, 8... After add /WX to compiter.

              L 1 Reply Last reply
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              • Y yu jian

                There is a error that Visual Studio 2008 only supports 1, 2, 4, 8... After add /WX to compiter.

                L Offline
                L Offline
                Lost User
                wrote on last edited by
                #17

                See the documentation[^], which clearly states that the only valid values for n are 1, 2, 4, 8 and 16. Thus using 0 is an invalid #pragma and will be ignored: the default packing (8) will be used.

                Binding 100,000 items to a list box can be just silly regardless of what pattern you are following. Jeremy Likness

                Y 1 Reply Last reply
                0
                • C Chris Losinger

                  Erudite_Eric wrote:

                  pragma pack 0 means there is no packing between data members, so they are contiguous in memory.

                  no. that's pack(1) : align on single bytes. pack(n) specifies the structure alignment, not the number of bytes between structs.

                  image processing toolkits | batch image processing

                  L Offline
                  L Offline
                  Lost User
                  wrote on last edited by
                  #18

                  Doh! Quite correct. :doh:

                  ============================== Nothing to say.

                  1 Reply Last reply
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                  • L Lakamraju Raghuram

                    I am reading 8 on my console.

                    L Offline
                    L Offline
                    Lost User
                    wrote on last edited by
                    #19

                    Yeah, it is 8, DWORD alignment. I got it wrong... :(

                    ============================== Nothing to say.

                    1 Reply Last reply
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                    • L Lost User

                      See the documentation[^], which clearly states that the only valid values for n are 1, 2, 4, 8 and 16. Thus using 0 is an invalid #pragma and will be ignored: the default packing (8) will be used.

                      Binding 100,000 items to a list box can be just silly regardless of what pattern you are following. Jeremy Likness

                      Y Offline
                      Y Offline
                      yu jian
                      wrote on last edited by
                      #20

                      Thank you for everyone's replies.

                      1 Reply Last reply
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                      • Y yu jian

                        The msdn says #pragma pack(n) will change current alignment value to n. But if n is zero, what will do?

                        A Offline
                        A Offline
                        Aescleal
                        wrote on last edited by
                        #21

                        According to the C++ standard it can do whatever it likes. It's a way of implementors switching on non-standard features of the compiler. I think Griff and the others have told you enough about what it does on VC++ though! Cheers, Ash PS: Except this is the one thread Griff hasn't posted in. Let's try "Chris and the others..." instead!

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