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Re: arithmetic operator precedence
- X-seq: zsh-workers 25184
- From: Stephane Chazelas <Stephane_Chazelas@xxxxxxxx>
- To: Zsh hackers list <zsh-workers@xxxxxxxxxx>, Richard Hartmann <richih.mailinglist@xxxxxxxxx>, Peter Stephenson <p.w.stephenson@xxxxxxxxxxxx>
- Subject: Re: arithmetic operator precedence
- Date: Tue, 17 Jun 2008 12:57:42 +0100
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On Tue, Jun 17, 2008 at 01:19:34PM +0200, Vincent Lefevre wrote:
> On 2008-06-17 10:45:09 +0100, Stephane Chazelas wrote:
> > But ** is not ^, it's a binary operator whose shape reminds that
> > of multiply, like a multiply++.
> So, why is ** right-associative while * is left-associative?
Don't know. I can't think of a good reason why 2**3**4 shoud be
2**(3**4) rather than (2**3)**4
4 / 4\ 4
3 \3 / rather than / 3\
2 2 \2 /
> > And even then, POSIX's ^ in bc is handled as -3^2 = 9.
> But note that bc is the only calculator with such an unfortunate
> choice. And I doubt that bc has been designed by end users. Also,
> perhaps those who wrote bc in the first place didn't think about
> this problem and just wanted to privilege the precedence of
> unary operators as it is often the case.
Which makes sense to me. I was serious when I said that -3 ** 2
was more intuitive to me. Because -3 for me looks like a single
BTW, I think I've got a rationale for -3 ** 2 == 9:
I think POSIX allows $((a * 3)) only as far as $a contains a
constant. If it contains "1 + 1", you won't be guaranteed to
have either 6 or 4.
SUSv3> If the shell variable x contains a value that forms a valid
SUSv3> integer constant, then the arithmetic expansions "$((x))" and
SUSv3> "$(($x))" shall return the same value.
So, if you agree that -3 is an integer constant, and agree that
as POSIX says variable expansion should be performed before the
arithmetic expression is evaluated, so that x=3; echo $(($x *
2)) is the same as echo $((-3 ** 2)), that'd mean tha
x=-3; $((x ** 2)) would either have to expand to -9, or $((-3 **
2)) should expand to 9.
> > It could be a good idea to ask ksh, POSIX/bc and perl authors for
> > the rationale behind their choices.
> I think that Perl authors would say something like conventional math
> writing (that's what some of authors of calculators say and what users
> often demand).
But I don't write 2**3 in conventional math writing, nor do I
I can understand there be different rules, because of the
linear, one character right to the previous one contraint of
computer strings while we have 2D freedom in hand writing.
1 + 2
has to be translated to (1+2)/3 because of that constraint.
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