Mail Archives: djgpp/1997/04/30/19:39:57
rellwood (rellwood AT aludra DOT usc DOT edu) wrote:
: On 30 Apr 1997, George Foot wrote:
: > rellwood (rellwood AT aludra DOT usc DOT edu) wrote:
: >
: > : The best method is the slowest, and that is to AND each nonzero pixel in
: > : sprite A with each nonzero pixel in sprite B. If any of the ANDs return
: >
: > I think you mean OR.
: No I meant AND. If OR was used, the function would return a true if it
: encountered a pixel from sprite A *OR* sprite B in a given location, which
: is not necesarially a collision. If AND was used it would only return
: true if it encountered a pixel from sprite A *AND* sprite B in a given
: location, which always signifies a collision between the two sprites.
Oops, my (partial) mistake. My line of thinking was that if pixelA==1 and
pixelB==2 then (pixelA&pixelB)==0; what I should have said was logical AND
not bitwise AND, i.e. (pixelA&&pixelB)==TRUE in this case.
--
George Foot <mert0407 AT sable DOT ox DOT ac DOT uk>
Merton College, Oxford
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