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  4. c libraries needed to deal with bitmaps

c libraries needed to deal with bitmaps

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  • C Calin Negru

    I`m having trouble understanding how this code works. The problem is I`m not a c expert.

    cf.Read( &BFH, sizeof( BITMAPFILEHEADER )

    For instance how does the portion of data read from my file fits into a structure. It`s not a structure of identical items BITMAPFILEHEADER is a mix of WORDs and DWORDS

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    Lost User
    wrote on last edited by
    #21

    The sizeof operator gives the number of bytes in the object referred to. So in your case it will be the size of a BITMAPFILEHEADER, and the read operation will read the correct number of bytes into the structure. When complete each field of the structure should be correctly aligned*. *There are conditions where this may not be true, so beware if you have a structure containing single byte elements. You may ignore that for the moment.

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    • L Lost User

      The sizeof operator gives the number of bytes in the object referred to. So in your case it will be the size of a BITMAPFILEHEADER, and the read operation will read the correct number of bytes into the structure. When complete each field of the structure should be correctly aligned*. *There are conditions where this may not be true, so beware if you have a structure containing single byte elements. You may ignore that for the moment.

      C Offline
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      Calin Negru
      wrote on last edited by
      #22

      thanks that makes sense. In the code above do I need m_nPaletteEntries at all? Is used to calculate the offset at which individual pixels are stored?

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      • C Calin Negru

        thanks that makes sense. In the code above do I need m_nPaletteEntries at all? Is used to calculate the offset at which individual pixels are stored?

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        Lost User
        wrote on last edited by
        #23

        You need to examine the different items in the info table as described at BITMAPINFOHEADER structure (Windows) | Microsoft Docs[^]. The values of the items related to colours will tell you whether a colour table is included or not.

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        • L Lost User

          You need to examine the different items in the info table as described at BITMAPINFOHEADER structure (Windows) | Microsoft Docs[^]. The values of the items related to colours will tell you whether a colour table is included or not.

          C Offline
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          Calin Negru
          wrote on last edited by
          #24

          I have to take a short break. Your insight (last post included) is priceless Richard, thanks

          `DreamLand Page` is my projects facebook page.

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          • C Calin Negru

            I have to take a short break. Your insight (last post included) is priceless Richard, thanks

            `DreamLand Page` is my projects facebook page.

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            Lost User
            wrote on last edited by
            #25

            Sorry to disappoint you, but my 'insight' is actually just a matter of reading the documentation.

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            • L Lost User

              Sorry to disappoint you, but my 'insight' is actually just a matter of reading the documentation.

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              Calin Negru
              wrote on last edited by
              #26

              the code you quoted is the bit that throws me into fog. What does the line that has the overload operator do?

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              • C Calin Negru

                the code you quoted is the bit that throws me into fog. What does the line that has the overload operator do?

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                Lost User
                wrote on last edited by
                #27

                fearless_ wrote:

                What does the line that has the overload operator do?

                What line?

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                • L Lost User

                  fearless_ wrote:

                  What does the line that has the overload operator do?

                  What line?

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                  Calin Negru
                  wrote on last edited by
                  #28

                  m_nPaletteEntries = 1 << m_pBIH->biBitCount;

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                  • C Calin Negru

                    m_nPaletteEntries = 1 << m_pBIH->biBitCount;

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                    phil o
                    wrote on last edited by
                    #29

                    This is a bitwise left-shift operation which is equivalent to

                    m_nPaletteEntries = pow(2, m_pBIH->biBitCount); // 2 to the power of bpp

                    What are shift operators in C++?[^] This trick is used quite often, since a bitwise shift operation is way quicker than the pow() method, which is rather intense on processing time. Edit: here's an interesting list of useful bitwise tricks: Bit Twiddling Hacks[^]

                    "Five fruits and vegetables a day? What a joke! Personally, after the third watermelon, I'm full."

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                    • P phil o

                      This is a bitwise left-shift operation which is equivalent to

                      m_nPaletteEntries = pow(2, m_pBIH->biBitCount); // 2 to the power of bpp

                      What are shift operators in C++?[^] This trick is used quite often, since a bitwise shift operation is way quicker than the pow() method, which is rather intense on processing time. Edit: here's an interesting list of useful bitwise tricks: Bit Twiddling Hacks[^]

                      "Five fruits and vegetables a day? What a joke! Personally, after the third watermelon, I'm full."

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                      Lost User
                      wrote on last edited by
                      #30

                      It's not pow, it's multiply by 2.

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                      • L Lost User

                        It's not pow, it's multiply by 2.

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                        phil o
                        wrote on last edited by
                        #31

                        For a left-shift by a single bit, yes, a multiplication by 2. But a left-shift by n bits, in the end, is the same as a multiplication by 2 to the power of n; since original value is 1, this leads to 2 to the power of n. Or am I missing something?

                        "Five fruits and vegetables a day? What a joke! Personally, after the third watermelon, I'm full."

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                        • P phil o

                          For a left-shift by a single bit, yes, a multiplication by 2. But a left-shift by n bits, in the end, is the same as a multiplication by 2 to the power of n; since original value is 1, this leads to 2 to the power of n. Or am I missing something?

                          "Five fruits and vegetables a day? What a joke! Personally, after the third watermelon, I'm full."

                          L Offline
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                          Lost User
                          wrote on last edited by
                          #32

                          I found myself thinking about this when I woke up this morning and you are, of course, quite correct.

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