每个像素的C ++位图位

Question

I'm trying to understand building a bmp based on raw data in c++ and I have a few questions.

My bmp can be black and white so I figured that the in the bit per pixel field I should go with 1. However in a lot of guides I see the padding field adds the number of bits to keep 32 bit alignment, meaning my bmp will be the same file size as a 24 bit per pixel bmp.

Is this understanding correct or in some way is the 1 bit per pixel smaller than 24, 32 etc?

Thanks

Answer 1

Monochrome bitmaps are aligned too, but they will not take as much space as 24/32-bpp ones.

• A row of 5-pixel wide 24-bit bitmap will take 16 bytes: 5*3=15 for pixels, and 1 byte of padding.
• A row of 5-pixel wide 32-bit bitmap will take 20 bytes: 5*4=20 for pixels, no need for padding.
• A row of 5-pixel wide monochrome bitmap will take 4 bytes: 1 byte for pixels (it is not possible to use less than a byte, so whole byte is taken but 3 of its 8 bits are not used), and 3 bytes of padding.

So, monochrome bitmap will of course be smaller than 24-bit one.

Answer 2

The answer is already given above (that bitmap rows are aligned/padded to 32-bit boundary), however if you want more information, you might want to read DIBs and Their Uses , the "DIB Header" section - it explains in detail.

Every scanline is DWORD-aligned. The scanline is buffered to alignment; the buffering is not necessarily 0.

The scanlines are stored upside down, with the first scan (scan 0) in memory being the bottommost scan in the image. (See Figure 1.) This is another artifact of Presentation Manager compatibility. GDI automatically inverts the image during the Set and Get operations. Figure 1. (Embedded image showing memory and screen representations.)