c libraries needed to deal with bitmaps
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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."