dumb question about parameter expansion

dumb question about parameter expansion

[sergio]

Hello.

Looks like it's dumb question, but I can't understand this.

% zsh -f
% l='ls'
% o='-l -h'
% $l $o
ls: invalid option -- ' '
Try `ls --help' for more information.

% ls -l -h
total 512
drwxr-x---  2 sergio sergio ...

-- 
sergio.

Re: dumb question about parameter expansion

[Jérémie Roquet]

Hi Sergio,

2012/10/11 sergio <mailbox@sergio.spb.ru>:
> Looks like it's dumb question, but I can't understand this.
>
> % zsh -f
> % l='ls'
> % o='-l -h'
> % $l $o
> ls: invalid option -- ' '
> Try `ls --help' for more information.
>
> % ls -l -h
> total 512
> drwxr-x---  2 sergio sergio ...

Unlike other shells, zsh doesn't split unquoted variables by default.
If you want that behaviour, you can either:
 - “setopt shwordsplit” to have it by default, or
 - use “$l $=o” to have it selectively (the “=” forces the split).

Best regards,

-- 
Jérémie

Re: dumb question about parameter expansion

[Frank Terbeck]

sergio wrote:
> Looks like it's dumb question, but I can't understand this.
>
> % zsh -f
> % l='ls'
> % o='-l -h'
> % $l $o
> ls: invalid option -- ' '
> Try `ls --help' for more information.
>
> % ls -l -h
> total 512
> drwxr-x---  2 sergio sergio ...

Zsh doesn't do wordsplitting of unquoted parameters (at least not by
default, which is a good thing[tm]). Use an array instead:

% o=(-l -h)
% ls $o

See <http://zshwiki.org/home/scripting/paramexp> as well.

Regards, Frank

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