Re: arithmetic operator precedence

[Stephane Chazelas <Stephane_Chazelas@xxxxxxxx>]

On Thu, Jun 19, 2008 at 06:10:56PM +0200, Mikael Magnusson wrote:
> 2008/6/19 Vincent Lefevre <vincent@xxxxxxxxxx>:
> > On 2008-06-19 14:29:03 +0200, Richard Hartmann wrote:
> >> On Thu, Jun 19, 2008 at 11:58, Peter Stephenson <pws@xxxxxxx> wrote:
> >> > That seems a good idea in any case.
> >>
> >> Can you add that this is done to remain both backwards and shell-
> >> compatible and that this is not what a 'real' language/calculator
> >> will do?
> >
> > This isn't shell-compatible. AFAIK, only bash and ksh93 support **.
> >
> >> Perhaps even include a reference to the POSIX argument.
> >
> > There's no POSIX argument. ** is just an extension.
> 
> I've followed this discussion, and I'm left wondering only one thing.
> If it's decided that $(( -1 ** 2 )) == 1, then what's the point? Why
> would anyone ever write that? Why require everyone to write -(1**2) when
> the expression makes no sense at all without the parentheses? Ie you would
> just write 1**2 without the - at all, nobody would ever write (-1)**2.
[...]

Think of:

double() {
  REPLY=$(($1 * 2))
}

square() {
  REPLY=$(($1 ** 2))
}

Where those functions are meant to be passed a number, positive
or negative, not any random arithmetic expression.

If you make ** have higher precedence than unary "-", then
square -1

returns -1

while double -1 returns -2 as expected.

-- 
Stéphane