Question #260626 on Yade changed: https://answers.launchpad.net/yade/+question/260626
Jan Stránský proposed the following answer: Hello, you can also easily try it yourself: import random; r=random.random O.bodies.append(sphere((r(),r(),r()),1)) base = '/tmp/a.' for ext in 'yade','xml': print O.bodies[0].state.pos[0] O.save(base+ext) O.reset() O.load(base+ext) print O.bodies[0].state.pos[0] my output: 0.872136848902 0.872136848902 0.872136848902 0.872136848902 it means that in both .yade and .xml format, the numbers (double precision numbers) are saved in maximum possible precision. Do you have any problems with O.save/O.load? cheers Jan 2015-01-21 9:41 GMT+01:00 Bruno Chareyre < [email protected]>: > Question #260626 on Yade changed: > https://answers.launchpad.net/yade/+question/260626 > > Bruno Chareyre proposed the following answer: > Hi, > Precision depends on which format you save to. If you save in binary > (filename.yade) it will have the maximum precision. > If you save in xml there is a conversion to string, and in that case I'm > not sure what happens. You could check that in boost::serialization. > I hope it helps. > Bruno > > -- > You received this question notification because you are a member of > yade-users, which is an answer contact for Yade. > > _______________________________________________ > Mailing list: https://launchpad.net/~yade-users > Post to : [email protected] > Unsubscribe : https://launchpad.net/~yade-users > More help : https://help.launchpad.net/ListHelp > -- You received this question notification because you are a member of yade-users, which is an answer contact for Yade. _______________________________________________ Mailing list: https://launchpad.net/~yade-users Post to : [email protected] Unsubscribe : https://launchpad.net/~yade-users More help : https://help.launchpad.net/ListHelp
