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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 was only using a part of the bitmap processing code I was using as reference. I don`t need the palette part so I trimmed out a bit too much.

    m_pDib = pDib;
    m_dwDibSize = dwDibSize;

    // Pointer our BITMAPINFOHEADER and RGBQUAD
    // variables to the correct place in the Dib data.
    m\_pBIH = (BITMAPINFOHEADER \*) m\_pDib;
    m\_pPalette =
    	(RGBQUAD \*) &m\_pDib\[sizeof(BITMAPINFOHEADER)\];
    
    //	get image width and height
    m\_width = m\_pBIH->biWidth;
    m\_height = m\_pBIH->biHeight;
    
    // Calculate the number of palette entries.
    m\_nPaletteEntries = 1 << m\_pBIH->biBitCount;
    if( m\_pBIH->biBitCount > 8 )
    	m\_nPaletteEntries = 0;
    else if( m\_pBIH->biClrUsed != 0 )
    	m\_nPaletteEntries = m\_pBIH->biClrUsed;
    
    // Point m\_pDibBits to the actual Dib bits data.
    m\_pDibBits =
    	&m\_pDib\[sizeof(BITMAPINFOHEADER)+
    		m\_nPaletteEntries\*sizeof(RGBQUAD)\];
    
    // If we have a valid palette, delete it.
    if( m\_Palette.GetSafeHandle() != NULL )
    	m\_Palette.DeleteObject();
    

    I know I must go past the bitmapinfoheader to get to the pixels

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

    Please use proper <pre> tags around code blocks (and proper indentation), not Quoted text. The following code sample is preceded by <pre lang="c++"> and followed by </pre>

    // Calculate the number of palette entries.
    m_nPaletteEntries = 1 << m_pBIH->biBitCount;
    if( m_pBIH->biBitCount > 8 )
    m_nPaletteEntries = 0;
    else if( m_pBIH->biClrUsed != 0 )
    m_nPaletteEntries = m_pBIH->biClrUsed;

    Much more readable.

    C 1 Reply Last reply
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    • L Lost User

      Please use proper <pre> tags around code blocks (and proper indentation), not Quoted text. The following code sample is preceded by <pre lang="c++"> and followed by </pre>

      // Calculate the number of palette entries.
      m_nPaletteEntries = 1 << m_pBIH->biBitCount;
      if( m_pBIH->biBitCount > 8 )
      m_nPaletteEntries = 0;
      else if( m_pBIH->biClrUsed != 0 )
      m_nPaletteEntries = m_pBIH->biClrUsed;

      Much more readable.

      C Offline
      C Offline
      Calin Negru
      wrote on last edited by
      #20

      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

      L 1 Reply Last reply
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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

        L Offline
        L Offline
        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.

        C 1 Reply Last reply
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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
          C Offline
          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?

          L 1 Reply Last reply
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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?

            L Offline
            L Offline
            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.

            C 1 Reply Last reply
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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
              C Offline
              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.

              L 1 Reply Last reply
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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.

                L Offline
                L Offline
                Lost User
                wrote on last edited by
                #25

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

                C 1 Reply Last reply
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                • L Lost User

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

                  C Offline
                  C Offline
                  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?

                  L 1 Reply Last reply
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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?

                    L Offline
                    L Offline
                    Lost User
                    wrote on last edited by
                    #27

                    fearless_ wrote:

                    What does the line that has the overload operator do?

                    What line?

                    C 1 Reply Last reply
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                    • L Lost User

                      fearless_ wrote:

                      What does the line that has the overload operator do?

                      What line?

                      C Offline
                      C Offline
                      Calin Negru
                      wrote on last edited by
                      #28

                      m_nPaletteEntries = 1 << m_pBIH->biBitCount;

                      P 1 Reply Last reply
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                      • C Calin Negru

                        m_nPaletteEntries = 1 << m_pBIH->biBitCount;

                        P Offline
                        P Offline
                        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."

                        L 1 Reply Last reply
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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."

                          L Offline
                          L Offline
                          Lost User
                          wrote on last edited by
                          #30

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

                          P 1 Reply Last reply
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                          • L Lost User

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

                            P Offline
                            P Offline
                            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."

                            L 1 Reply Last reply
                            0
                            • 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
                              L Offline
                              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.

                              1 Reply Last reply
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