c libraries needed to deal with bitmaps
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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.
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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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
The
sizeofoperator gives the number of bytes in the object referred to. So in your case it will be the size of aBITMAPFILEHEADER, 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. -
The
sizeofoperator gives the number of bytes in the object referred to. So in your case it will be the size of aBITMAPFILEHEADER, 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.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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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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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.
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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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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Sorry to disappoint you, but my 'insight' is actually just a matter of reading the documentation.
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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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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m_nPaletteEntries = 1 << m_pBIH->biBitCount;
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m_nPaletteEntries = 1 << m_pBIH->biBitCount;
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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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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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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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."