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 <msp@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, apvague@gmail.com wrote:
>
>
> Sent from my iPhone
>
> > On May 31, 2016, at 4:44 PM, Raphaël Ilias <phae.ilias@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
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