Re: forcing float arithmetic.

[Daniel Shahaf <d.s@xxxxxxxxxxxxxxxxxx>]

Ray Andrews wrote on Wed, Mar 24, 2021 at 18:12:29 -0700:
> On 2021-03-24 2:59 p.m., Bart Schaefer wrote:
> > On Wed, Mar 24, 2021 at 2:54 PM Roman Perepelitsa
> > <roman.perepelitsa@xxxxxxxxx> wrote:
> > > The behavior of zsh in this regard is consistent with C and all languages inspired by it (C++, Java, C# and many, many others). This is really working as intended.
> > Proof:
> > 
> > #include <stdio.h>
> > void main() {
> >   int x = 3, y = 7;
> >   float z = (x/y);
> >   printf("%g\n", z);
> > }
> > 
> Long time since I did any floating point in C, so I'll take you guy's word
> for it.  Final shot would that since one can force the conversion by, say,
> multiplying by 1.0, which is otherwise pointless, one could imagine some
> option whereby the bother is simply not required.
> 
> (( aa = ((2 * nn) - 1) / (nn**2.0) ))
> 
> ... in that case the denominator is not changed in any way, yet if flags
> that the division should be passed as a float.  Seems an awkward way of
> getting the conversion.  How is 2^2 different from 2^2.0 ?  If 'aa' was an
> integer then of course the result must be rounded, but it still seems to me
> the float should receive the actual result.

The «/» operator is defined to perform integer division (discarding the
remainder) when both of its arguments are integers, and floating-point
division otherwise.

That _is_ inconsistent, in a way, since in C it's not possible to
implement a two-argument function that has the same semantics as the «/»
operator for both integral and floating-point types… but that's how it is.

> One of my little whines, nothing of substance. Still one might dream
> of setopt AUTO_FLOAT.

Python 3's division operator always returns floats.

Daniel