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**Subject**: **Re: [ANN] Lua 5.4.0 (work2) now available**
**From**: Roberto Ierusalimschy <roberto@...>
**Date**: Fri, 24 Aug 2018 17:12:23 -0300

> > Lua does use fmod but it doesn't have exactly the same semantics. See the comment on luai_nummod on limits.h for details.
>
> More exactly, Lua uses this definition:
>
> { (m) = l_mathop(fmod)(a,b); if ((m)*(b) < 0) (m) += (b); }
>
> The correction is due to fmod assumes a rounding towards zero in
> the division, while Lua rounds towards minus infinite. But the test
> adds little to the cost; 'fmod' is the expensive operation here.
>
> Earlier Lua versions used the mathematical definition:
>
> (m) = ((a) - l_mathop(floor)((a)/(b))*(b))
>
> It is *much* faster than 'fmod', but it gives NaN when 'b' is infinite
> (it should give 'a'). I am not sure whether there are other problematic
> cases besides 'b' being inf or -inf.
>
> -- Roberto
Apparently, the following formula gives the correct answer for all
cases:
{ (m) = l_mathop(floor)((a)/(b)); \
(m) = ((m) == 0) ? ((a) * (b) < 0 ? (a) + (b) : (a)) : (a) - (m)*(b); }
In my machine, it can be up to 5x faster than 'fmod'. ('fmod' time
seems to depend heavily on the parameters.)
-- Roberto