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Re: Asking ZSH: How are you ?
- X-seq: zsh-users 9070
- From: DervishD <zsh@xxxxxxxxxxxx>
- To: Meino Christian Cramer <Meino.Cramer@xxxxxx>
- Subject: Re: Asking ZSH: How are you ?
- Date: Sat, 9 Jul 2005 17:17:36 +0200
- Cc: zsh-users@xxxxxxxxxx
- In-reply-to: <20050709.165213.74740823.Meino.Cramer@xxxxxx>
- Mail-followup-to: Meino Christian Cramer <Meino.Cramer@xxxxxx>, zsh-users@xxxxxxxxxx
- Mailing-list: contact zsh-users-help@xxxxxxxxxx; run by ezmlm
- Organization: DervishD
- References: <20050709.165213.74740823.Meino.Cramer@xxxxxx>
Hi Meino :)
* Meino Christian Cramer <Meino.Cramer@xxxxxx> dixit:
> When listing options (*** by `setopt', `unsetopt', `set -o' or
> `set +o'***), those turned on by default appear in the list
> prefixed with `no'. Hence (unless KSH_OPTION_PRINT is set),
> `setopt' shows all options whose settings are changed from the
> default.
>
> but giving "setopt +o" or "setopt -o"
> always gives me (even for setopt +o!):
>
> setopt: string expected after -o
Of course it does, it's correct ;) You're mixing 'set +o/-o' with
'setopt' and 'unsetopt'. If you want to list all options use
'setopt', without arguments, or 'set -o'. If you want to show ALL
options, no matter if they have their default value or not, use this
little snippet (for example):
for option in ${(ko)options}; print ${(r:21:}option $options[$option]
This will give you the list of all option names followed by its
value. I find this easier to understand that the default of prefixing
with 'no' and the like.
Raúl Núñez de Arenas Coronado
--
Linux Registered User 88736 | http://www.dervishd.net
http://www.pleyades.net & http://www.gotesdelluna.net
It's my PC and I'll cry if I want to...
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