Re: [Bug] math.mod behaves differently than the modulus operator.

[Daniel Stephens <daniel@...>]

The manual doesn't claim that % and math.mod are the same, it simply says:

a % b == a - math.floor(a/b)*b


And in this case that IS consistent with your observed results.



Polarina wrote:
> math.mod behaves differently than the modulus operator.
>
> How to reproduce:
>   
> > = math.mod(7, -(1 / 0))
> >     
> 7
>   
> > = 7 % -(1 / 0)
> >     
> nan
>
>
>
>