Re: [Algorithms] OT: How to read a bitmap of arbitrary size?

[Pai-Hung C. <pa...@ac...>]

Hi,

That explains everything! :-)  Thanks a lot!  But could you explain what's
the meaning of ((bytes_per_pixel * pixels_per_row) + 3) & (~3), specifically
the ...+ 3) & (~3) part?

Thank you,

Pai-Hung Chen


> For Windows BMP files, the pitch must be a multiple of 4 bytes. 5 pixels *
3
> bytes = 15 bytes is not a multiple of 4! In such cases, the BMP format
uses
> the next higher value (that is multiple 4) as the pitch. In this specific
> case, the next higher value that is a multiple of 4, is 16. Thus the
Windows
> BMP format uses a pitch of 16 bytes.
> So for a 5x5 BMP you have 5 * (((3 * 5) + 3) & (~3)) = 5 * 16 = 80 bytes.
> (Note, that the additional byte is unused).
>
> In order to compute the pitch for a bit-count >= 8, you can use the
> following formula:
> pitch = ((bytes_per_pixel * pixels_per_row) + 3) & (~3)
>
> biSizeImage then is: biSizeImage = pitch * num_pixel_rows
>
> HTH,
> Niki
>
> ----- Original Message -----
> From: Pai-Hung Chen <pa...@ac...>
> To: <gda...@li...>
> Sent: Sunday, August 13, 2000 6:57 PM
> Subject: [Algorithms] OT: How to read a bitmap of arbitrary size?
>
>
> > Hi,
> >
> > I have been frustratingly bugged for hours and I decide to resort to the
> > list.  I want to read a 24-bit bitmap heightfield.  I can correctly read
> it
> > as long as the dimension of the bitmap file is in 2's power such as 4x4,
> > 8x8, 16x16, etc.  (But 2x2 doesn't work?!)  In these _normal_ cases, the
> > *biSizeImage* field of the BITMAPINFOHEADER structure gives me the
correct
> > size of image in bytes.  However, if I create a bitmap of non-2's-power
> > dimension such as 5x5, 6x6, 7x7, etc., the *biSizeImage* field gives me
> some
> > bogus number.  For example, 5x5 yields 80 bytes (5x5x3 = 75 bytes); 6x6
> > yields 120 bytes (6x6x3 = 108 bytes); 7x7 yields 168 bytes (7x7x3 = 147
> > bytes).  Could somebody tell me what happened?
> >
> > Thanks in advance,
> >
> > Pai-Hung Chen
> >
> >
> >
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