Re: zshexpn problem

[Peter Stephenson <pws@csr.com>]

Matthias Berndt wrote:
> On Sun, 20 Feb 2005 05:33:35 +0100
> Matthias Berndt <Berndt.Matthias@gmx.de> wrote:
> 
> > Hi,
> > 
> > When LVERSION=2.6.10 ... why:
> > print ${(s/./)LVERSION[2]}
> > ->

The s splitting happens late; it means the overall result of the
expansion is an array.  It doesn't mean the value is instantly turned
into an array.  (It's not evaluated simply left to right, for example.)

There are a list of rules in the manual page for the order in which an
expansion takes place at each level of nesting, and they have this order:

2. _Parameter Subscripting_
  i.e. the [2]
8. _Forced Splitting_
  i.e. the (s/./)

So actually it's not well defined what happens in this case.

Obviously, "documented" isn't the same as "clear".  The following sentence
is the key part of the documentation:

  Note that the Zsh Development Group accepts _no responsibility_ for any
  brain damage which may occur during the reading of the following rules.

> > print ${${(s/./)LVERSION}[2]}
> > -> 6
> > 
> > ... works 'correct', but is seems not very intelligent.

That's the correct, logical way to do it, in as much as zsh is ever
logical.  Generate an expression which returns an array, then apply the
subscript to that.  Then rule 1 comes into force.

1. _Nested Substitution_
     If multiple nested ${...} forms are present, substitution is
     performed from the inside outwards.  At each level, the
     substitution takes account of whether the current value is a
     scalar or an array, whether the whole substitution is in double
     quotes, and what flags are supplied to the current level of
     substitution, just as if the nested substitution were the
     outermost.
     ...

I still marvel that people complain about Perl being obscure.

-- 
Peter Stephenson <pws@csr.com>                  Software Engineer
CSR PLC, Churchill House, Cambridge Business Park, Cowley Road
Cambridge, CB4 0WZ, UK                          Tel: +44 (0)1223 692070


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