Re: Finding bitprecision

If that is the question, the answer is getlocalbitprec().

Best,
Aurel

On 19/12/2023 09:52, Loïc Grenié wrote:
> On Tue Dec, 19th, 2023 at 09:08, Aurel Page wrote:
>
>     This is the time to create 1. at your precision, not the time to find
>     the bitprecision.
>
>     ? my(s=0,p=10^4); localprec(p); for(n=1,10^5,a=1.; s+=p);
>     time = 74 ms.
>     ? my(s=0,p=10^6); localprec(p); for(n=1,10^5,a=1.; s+=p);
>     time = 2,718 ms.
>
>
>     Indeed, however I think Ilya asked why there is no way to know the
>   local bit precision without creating an object which require time to
>   create.
>
>      You can install(get_localbitprec,l), for a workaround:
>
> ? install(get_localprec,l)
> ? localbitprec(1000);get_localprec()
> %2 = 18
> ? localbitprec(10000);get_localprec()
> %3 = 159
>
>        Best,
>
>              Loïc
>
>     On 19/12/2023 07:20, Ilya Zakharevich wrote:
>     > Should not there be O(1) way to find the current
>     localbitprecision()?
>     >
>     >    (22:17) gp > localprec(10000);   my(s); for(n=1,100000, 
>     s+=bitprecision(1.));
>     >    time = 78 ms.
>     >    (22:17) gp > localprec(1000000); my(s); for(n=1,100000, 
>     s+=bitprecision(1.));
>     >    time = 2,121 ms.
>     >
>     > Thanks,
>     > Ilya
>
>