*(WORD*)((BYTE*)ddsd.lpSurface+y*ddsd.lPitch+x*2)
but i get something weird.
Please help me, someone.
Octavian
E-mail: octav@octavsoft.com
Webpage: http://www.octavsoft.com
need help with locking dd surf
Author
When I lock a dd surface, how can I get to pixel (x,y)??
Let''s say that I lock rectangle rc, and in ddsd I have the surface description. What''s the formula to get to the memory for pixel (x,y), relative to the upper-left corner of rc. The bpp is 16bits.
I tried:
Try this ...
This is all assuming that you locked the surface with the proper perameters such as DDLOCK_SURFACEMEMORYPTR in the flags when you lock the surface...
Also note that x and y are the co-ordinates on the screen - I hope this helps ...
Sincerely,
Destroyer
===============================
In a world without boundaries -
Who needs gates or windows ?
=============================== [/source]
Edited by - Destroyer on June 25, 2001 4:30:02 PM
WORD *pixbuffer = (WORD *)ddsd.lpSurfacepixbuffer[x+y*ddsd.lPitch]; This is all assuming that you locked the surface with the proper perameters such as DDLOCK_SURFACEMEMORYPTR in the flags when you lock the surface...
Also note that x and y are the co-ordinates on the screen - I hope this helps ...
Sincerely,
Destroyer
===============================
In a world without boundaries -
Who needs gates or windows ?
=============================== [/source]
Edited by - Destroyer on June 25, 2001 4:30:02 PM
In 16bpp mode if you are going to cast the buffer to a word* you will need to divide the pitch by 2. If you are going to cast to a byte* you need to mult x by 2:
WORD *pixbuffer = (WORD *)ddsd.lpSurface;
pixbuffer += (x + (y * (ddsd.lpitch/2)));
*pixbuffer = color;
or
BYTE *pixbuffer = (BYTE *)ddsd.lpSurface;
pixbuffer += ((x*2) + (y * ddsd.lpitch));
*pixbuffer = color;
Lots of room for optimization here, but you get the picture.
WORD *pixbuffer = (WORD *)ddsd.lpSurface;
pixbuffer += (x + (y * (ddsd.lpitch/2)));
*pixbuffer = color;
or
BYTE *pixbuffer = (BYTE *)ddsd.lpSurface;
pixbuffer += ((x*2) + (y * ddsd.lpitch));
*pixbuffer = color;
Lots of room for optimization here, but you get the picture.
Uh oh...
Thats the first I''ve heard about dividing or multipying by 2 - why do you have to do that ?
Destroyer
Thats the first I''ve heard about dividing or multipying by 2 - why do you have to do that ?
Destroyer
Author
I figured it out in the end. Aldacron is right, because each pixel is 2 bytes (1 word) in 16 bpp. he pitch is always in bytes, so the width has to be multiplied with 2 or the oder way around. The fastest one would be:
cause there is no divide or multiplication.
Octavian
E-mail: octav@octavsoft.com
Webpage: http://www.octavsoft.com
*( (WORD*) ( (BYTE*) surf + y * ddsd1.lPitch ) + x )cause there is no divide or multiplication.
Octavian
E-mail: octav@octavsoft.com
Webpage: http://www.octavsoft.com
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