SDL error when no SDL involved
Author
Hi, I have a strange problem that involves (so it seems) SDL, but the code where tha error occours does not use SDL
Well, I have a simple geometric library I've written for some ray tracing tests. In the ray/triangle intersection routine there is the following line:
real dist;
.
.
.
dist = -((r.GetOrigin() - v[0]) ^ normal) / (r.GetDir() ^ normal);
where r is the ray, v[0] is a vertex of the current polygon and normal is the normal of the polygon. I use SDL to output the rendering results. Right at that line the program crashes and the following line is written on the console:
Fatal signal:Segmentation Fault <SDL Parachute Deployed>
I use winXP and Mingw, latest versions of MingW and SDL. The previous code worked well some ago. Then I did many changes to the application testing it only with spheres. Then I tried the polygons and that happens. Sorry for the lack of informations, if you have an idea (or want to have more code) then just say it.
Thank you!
EDIT: I forget to say that if I comment out this line, the error does not occour!
i dont know either what that has to do with SDL, but i cant blame your compiler for failing to figure out that line. what does it do? the ^operator evidently returns a vector, otherwise you couldnt substract the result from origin, a vector also i assume.
but then you proceed to divide two vectors by eachother? storing the result in a real... that has me confused, maybe your compiler aswell?
but then you proceed to divide two vectors by eachother? storing the result in a real... that has me confused, maybe your compiler aswell?
Quote:Original post by Eelco
i dont know either what that has to do with SDL, but i cant blame your compiler for failing to figure out that line. what does it do? the ^operator evidently returns a vector, otherwise you couldnt substract the result from origin, a vector also i assume.
but then you proceed to divide two vectors by eachother? storing the result in a real... that has me confused, maybe your compiler aswell?
Agreed. The ^ operator is a bitwise XOR. If you are wanting to do powers, you must do something this:
dist = - pow(r.GetOrigin() - v[0],normal) / pow(r.GetDir(),normal);That's assuming r.GetOrigin() returns some simple data type as well as r.GetDir(). If you are returning vectors, you might need to do some operator overloading (of which you cannot overload ^) and handle it like that. You are getting a SDL error with non-sdl code because you are generating an illegal instruction in a SDL program, so SDL must quit as well as all of its subsystems.
[edit] So you can overload the ^ operator [lol]. Silly me.
[Edited by - Drew_Benton on March 8, 2005 1:30:05 PM]
SDL catches fatal exceptions like this one (which is an access violation). Your code is throwing it, but SDL's error routines are catching it.
Author
Yeah, sorry, I had to be more clear: the ^ operator is the dot product on vectors, so the line
dist = -((r.GetOrigin() - v[0]) ^ normal) / (r.GetDir() ^ normal);
has the form real = real / real;
(I overloaded the operator ^ as the dot product on vectors).
I've found that the error is elsewhere. Tomorrow I will do some tests and eventually I will post some more informations. Thank you aniway!
dist = -((r.GetOrigin() - v[0]) ^ normal) / (r.GetDir() ^ normal);
has the form real = real / real;
(I overloaded the operator ^ as the dot product on vectors).
I've found that the error is elsewhere. Tomorrow I will do some tests and eventually I will post some more informations. Thank you aniway!
my guess is that '(r.GetDir() ^ normal)' is be returning 0, and causing a divide-by-zero error.
use SDL_Init( .... SDL_INIT_NOPARACHUTE ) to get rid of this message, the app then may just crash :)
Author
Quote:
my guess is that '(r.GetDir() ^ normal)' is be returning 0, and causing a divide-by-zero error.
Yes, I thought that too, but then I checked it and it is never 0. I've solved the problem, it was just a null pointer not handled in the proper way in another function. This stupid error has wasted my time, but yours also, so sorry. Thank you anyway!
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