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

[Klaus H. <k_h...@os...>]

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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