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

[John S. <jse...@ho...>]

That part is simply rounding up to the nearest 4.

John Sensebe
jse...@ho...
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----- Original Message -----
From: "Pai-Hung Chen" <pa...@ac...>
To: <gda...@li...>
Sent: Sunday, August 13, 2000 1:31 PM
Subject: Re: [Algorithms] OT: How to read a bitmap of arbitrary size?


> 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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> > >
> >
> >
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>
>
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