Sorry my error, I tried : random 1e+08 / 1e+08
..and it works. Don't know how I got this wrong results. It evens work until 1e+09, the precision error (that outputs 0) starts from 1e+10. Is 1e+09 using the full decimal precision of pd's 32-bits floats ? thanks Raphaël 2016-05-31 17:00 GMT+02:00 Miller Puckette <m...@ucsd.edu>: > I just tried "random 1e8" and "/ 1e8" and it seems to work for me. If it > doens't for you that's a bug I should look at :) > > cheers > Miller > > On Tue, May 31, 2016 at 04:48:59PM +0200, apva...@gmail.com wrote: > > > > > > Sent from my iPhone > > > > > On May 31, 2016, at 4:44 PM, Raphaël Ilias <phae.il...@gmail.com> > wrote: > > > > > > Hello list, > > > > > > I want to do a simple task : pick a random float in a defined range > (let's say between 0.0 and 1.0). > > > I know there are this kind of objects in external libraries (something > like [randomF] if remember), but since the collapse of Pd-Extended, I > generally prefer to make vanilla abstractions. > > > > > > The two solutions i foresee are : > > > > > > solution #1 : > > > > > > [random 1e+06] > > > | > > > [/ 1e+06] > > > > > > but this way, it doesn't use the full floating-point resolution ?... > and going over that range (like 1e_07) will result in errors (outputs 0) I > guess because of floating-point complexity. > > > > > > > > > # solution #2 : > > > > > > [noise~] > > > | > > > [snapshot~] > > > > > > but this won't work if DSP/audio is turned off. > > > > > > > > > > > > so I wondered if there are other vanilla and efficient solutions? > > > ...or if I just have to get the external from deken... ? > > > :) > > > > > > thanks, > > > > > > Raphaël > > > _______________________________________________ > > > Pd-list@lists.iem.at mailing list > > > UNSUBSCRIBE and account-management -> > https://lists.puredata.info/listinfo/pd-list > > > > _______________________________________________ > > Pd-list@lists.iem.at mailing list > > UNSUBSCRIBE and account-management -> > https://lists.puredata.info/listinfo/pd-list >
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