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Index › Suggestions & Bugs › Website,  Documentation,  Book Bugs › exit() not behaving like reference says

TorfusPolymorphus YaBB Newbies Offline Posts: 12

exit() not behaving like reference says Nov 14th, 2007, 1:46am   I've got trouble understanding how exit() works. According to the reference, Quote: Rather than terminating immediately, exit() will cause the sketch to exit after draw() has completed. But the output of the following sketch Code: void setup() { size(100,100); println("Before exit()"); exit(); println("After exit()"); } void draw() { println("draw()"); line(0,0,width,height); } is Code: Before exit() After exit() So draw() is not called at all, and still exit() does not terminate the program immediately. What am I missing?

mflux Senior Member Offline Posts: 388	Re: exit() not behaving like reference says Reply #1 - Nov 14th, 2007, 2:23am   Hmmm you might not be able to call exit() during setup. I believe it's supposed to be explicitly used in draw() and may be victim to some documentation oversight. From what I understand exit() signals an exit call WITHIN draw() so that once draw finishes the program will stop and draw will cease to become looped.
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fry YaBB Moderator Offline Posts: 2623 mass, usa	Re: exit() not behaving like reference says Reply #2 - Nov 15th, 2007, 8:23pm   i'll change the reference to say "after draw() has completed (if called inside draw) or after setup() has completed (if called during setup)".

TorfusPolymorphus YaBB Newbies Offline Posts: 12	Re: exit() not behaving like reference says Reply #3 - Nov 21st, 2007, 8:15pm   Okay, thanks for straightening this out.