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Date: Thu, 9 Apr 2015 18:31:04 -0700
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Subject: Re: `[[ -n $VAR ]]` equal to `[[ $VAR ]]`?
From: Kurtis Rader <krader@skepticism.us>
To: Thorsten Kampe <thorsten@thorstenkampe.de>
Cc: Zsh Users <zsh-users@zsh.org>
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When I run the following

    [[ $VAR ]] && print yes

I get a parse error. Which is what I expected given the documentation in
section "Conditional Expressions" of "man zshall". Are you seeing different
behavior? What makes you think a bare variable is a valid expression?

On Thu, Apr 9, 2015 at 6:23 PM, Thorsten Kampe <thorsten@thorstenkampe.de>
wrote:

> Hi,
>
> this is kind of a follow-up to
> http://www.zsh.org/mla/users/2011/msg00284.html .
>
> Basically the solutions were `((+VAR))` for zsh and `[[ -n
> ${VAR+stuff} ]]` for bash and zsh.
>
> Now my question is, isn't `[[ -n $VAR ]]` equivalent to `[[ $VAR ]]`?
> (`[[ -n ${VAR+stuff} ]]` equivalent to `[[ ${VAR+stuff} ]]`)
>
> Thorsten
>
>


-- 
Kurtis Rader
Caretaker of the exceptional canines Junior and Hank

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