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Re: `[[ -n $VAR ]]` equal to `[[ $VAR ]]`?
- X-seq: zsh-users 20123
- From: Kurtis Rader <krader@xxxxxxxxxxxxx>
- To: Thorsten Kampe <thorsten@xxxxxxxxxxxxxxxx>
- Subject: Re: `[[ -n $VAR ]]` equal to `[[ $VAR ]]`?
- Date: Thu, 9 Apr 2015 19:05:59 -0700
- Cc: Zsh Users <zsh-users@xxxxxxx>
- In-reply-to: <CABx2=D9Mk01ovQkMyWT_1nh9WdDvcpqmMJAOdka-oEmN8Xe4nw@mail.gmail.com>
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Bart's reply explains why I get a parse error: Mac OS X still has zsh
v5.0.5 which does not support the bash semantics for a bare string inside
[[ ]]. Even if you don't need compatibility with pre v5.0.6 zsh releases I
would discourage that syntax because of its ambiguity.
On Thu, Apr 9, 2015 at 7:02 PM, Kurtis Rader <krader@xxxxxxxxxxxxx> wrote:
> Okay, I had forgotten that a bare string is equivalent to "-n string" in
> bash. The zsh documentation makes no mention of this "feature". It probably
> works for you and not me because of an option that differs between our two
> environments. Although for the life of me I can't figure out what that
> option is.
> On Thu, Apr 9, 2015 at 6:39 PM, Thorsten Kampe <thorsten@xxxxxxxxxxxxxxxx>
>> * Kurtis Rader (Thu, 9 Apr 2015 18:31:04 -0700)
>> > 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
>> > behavior? What makes you think a bare variable is a valid expression?
>> if [[ $VAR ]]
>> printf "something\n"
>> printf "nothing\n"
>> Works fine in zsh and bash.
>> Same goes for
>> `[[ $VAR ]] && printf "something\n" || printf "nothing\n"`
> Kurtis Rader
> Caretaker of the exceptional canines Junior and Hank
Caretaker of the exceptional canines Junior and Hank
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