On Fri, 29 Mar 2024, Jordan LeDoux wrote:
> On Wed, Mar 27, 2024 at 12:08 AM Aleksander Machniak <a...@alec.pl> wrote:
>
> > On 27.03.2024 01:03, Saki Takamachi wrote:
> > >> $num = new BcNum('1.23', 2);
> > >> $result = $num + '1.23456';
> > >> $result->value; // '2.46456'
> > >> $result->scale; // ??
> > >
> > > In this case, `$result->scale` will be `'5'`. I added this to the
> > > RFC.
> >
> > I'm not sure I like this. Maybe we should be more strict here and
> > treat the $scale in constructor (and later withScale()) as the
> > actual scale for all operations.
> >
> >
> For addition, it absolutely should expand scale like this, unless the
> constructor also defines a default rounding type that is used in that
> situation. All numbers, while arbitrary, will be finite, so addition
> will always be exact and known based on inputs prior to calculation.
>
> Treating scale like this isn't more strict, it's confusing. For
> instance:
>
> ```
> $numA = new Number('1.23', 2);
> $numB = new Number('1.23456', 5);
>
> $expandedScale1 = $numA + $numB; // 2.46456
> $expandedScale2 = $numB + $numA; // 2.46456
>
> $strictScale1 = $numA + $numB; // 2.46 assuming truncation
> $strictScale2 = $numB + $numA; // 2.46456
> ```
>
> I ran into this same issue with operand ordering when I was writing my
> operator overload RFC.
>
> There are ways you could do the overload implementation that would get
> around this for object + object operations, but it's also
> mathematically unsound and probably unexpected for anyone who is going
> to the trouble of using an arbitrary precision library.
>
> Addition and subtraction should automatically use the largest scale
> from all operands. Division and multiplication should require a
> specified scale.
I agree. I think add/subtract also should always take the largest scale
here.
cheers,
Derick
