From: mert0407 AT sable DOT ox DOT ac DOT uk (George Foot) Newsgroups: comp.os.msdos.djgpp Subject: Re: Collision Detection Date: 30 Apr 1997 19:03:23 GMT Organization: Oxford University, England Lines: 23 Message-ID: <5k851r$6hc@news.ox.ac.uk> References: <5k75ea$j15 AT news DOT ox DOT ac DOT uk> NNTP-Posting-Host: sable.ox.ac.uk To: djgpp AT delorie DOT com DJ-Gateway: from newsgroup comp.os.msdos.djgpp Precedence: bulk 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 Merton College, Oxford