Subject: Re: bug#20443: grep pattern is forward slash + star

From: Eric Blake <[email-masked]>
Date: Mon, 27 Apr 2015 14:53:02 -0600

tag 20443 notabug
thanks

On 04/27/2015 01:05 PM, Clay Hansen wrote:
> Dear Coders,
> 
> if I echo 'a  *  z' | grep '/*' it succeeds and prints 'a  *  z'.
> Can this be right?

Yes.  The regular expression '/*' says to match 0-or-more instances of
'/'. 'a  *  z' has zero instances of '/', so it matches and gets printed.

> 
> echo 'a  *  z' | grep -F '/*' fails as expected.

That's the literal pattern '/*', which does not appear in your input.

Remember, to convert the literal pattern of '/*' under -F to a regular
expression when not using -F, you must escape any characters that are
otherwise special to regular expressions.  Your second command would be
equivalent to:

echo 'a  *  z' | grep '/\*'

where the '*' now matches a literal star rather than being the
0-or-more-operator.

As such, this is not a bug, so I'm closing the report.  However, feel
free to ask further questions on the topic.

-- 
Eric Blake   eblake redhat com    +1-919-301-3266
Libvirt virtualization library http://libvirt.org