Re: [Help-bash] Undocumented caret logical operator?

[Eric Blake]

On 05/02/2018 01:09 PM, G. Branden Robinson wrote:
> Here's why I asked.  There's a piece of OpenSSH that seems to believe ^
> is, or might be, an operator.  It's down in the weeds of some
> shell-detection madness, so I'm not sure if they're on crack or not.
>
> https://pastebin.com/df7jYEib

(Pastebins expire, but the mailing list archives do not; copying some of 
that example here)

# check that we have something mildly sane as our shell, or try to find 
something better
if false ^ printf "%s: WARNING: ancient shell, hunting for a more modern 
one... " "$0"
then

Yes, on all modern systems (and any POSIXy shell), this executes the 
single command 'false' with ignored arguments, and skips the 'then' 
clause.  Meanwhile, on old Solaris systems, where /bin/sh is Bourne 
shell rather than POSIX sh and has a ^ operator, this is parsed the same 
as 'false | printf ...', which executes the printf command (which may or 
may not have been a builtin in that old shell, and may or may not be on 
your PATH); it is HOPING that the exit status of printf will be 0, and 
that the WARNING message was printed, and that the 'then' clause then 
gets a chance to try re-execing under a saner shell.

But since printf is a relatively modern shell invention, I'm seriously 
wondering if a system old enough to have /bin/sh that understands ^ as a 
pipe operator would also lack printf, and thus break this snippet.

--
Eric Blake, Principal Software Engineer
Red Hat, Inc.           +1-919-301-3266
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