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Re: how to force scalar to be an array?
- X-seq: zsh-users 7943
- From: Bart Schaefer <schaefer@xxxxxxxxxxxxxxxx>
- To: ZSH User List <zsh-users@xxxxxxxxxx>
- Subject: Re: how to force scalar to be an array?
- Date: Sat, 28 Aug 2004 11:08:52 -0700 (PDT)
- In-reply-to: <20040828074355.GA4953@xxxxxxxxx>
- Mailing-list: contact zsh-users-help@xxxxxxxxxx; run by ezmlm
- References: <20040827154905.GA26846@xxxxxxxxx> <Pine.LNX.4.61.0408271416370.4189@xxxxxxxxxxxxxxxxxx> <20040828074355.GA4953@xxxxxxxxx>
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On Sat, 28 Aug 2004, Andy Spiegl wrote:
> > So your only safe bet is to actually assign the scalar to an array, and
> > then use subscripting on the array.
> But then how would I be able to split the words if there are more than one?
What I meant was:
words=( $=result )
You can even do it in one expression:
Note that you have to put the splitting flag on the LEFT side of ::= or
you don't get the array behavior from the subscript when there is only one
element. That works with the (s) parameter flag and its relatives, too;
> Hm, I think I must be doing something wrong here. Shouldn't it be an
> absolutely easy and common task to split a line into its words and then
> pick the first one and the rest of them.
Yes, you'd think so, but it's been zsh's behavior for a long time to treat
a one-element array as a scalar in certain contexts. You'd also think
that the (@) flag would fix that, but it only *preserves* array-ness in
the transition from inner to outer nested expansions, it doesn't *create*
array-ness from a scalar.
The latter might be something we could consider changing without much risk
of breaking anything.
> In Perl I'd do for example: (just to illustrate what I want to do)
> @words = split (/\s/, $result);
> or maybe:
> $result =~ /(\S+)(.*)/;
You can do that latter one in zsh, too, if you have extendedglob set:
print -l $match $match
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