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Subject: Re: 10.2-RELEASE/amd64: grep regex syntax vs. grep-bug
To: "no@spam@mgEDV.net" <nospam@mgedv.net>, freebsd-questions@freebsd.org
References: <00e801d0f9e2$125ec430$371c4c90$@mgedv.net>
From: "Chad J. Milios" <milios@ccsys.com>
Message-ID: <56093CB4.6060606@ccsys.com>
Date: Mon, 28 Sep 2015 09:12:20 -0400
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On 9/28/2015 7:37 AM, no@spam@mgEDV.net wrote:
> hi folks,
> the goal: grep, that a variable contains a number and JUST digits.
> # setup sample variable - we're on /bin/sh @ 10.2-RELEASE/amd64.
> x=""
> # od output the content of the variable to ensure content
> echo "$x" | od -ctdC
> 0000000   \n
>             10
> 0000001
> # now try /usr/bin/grep'pin (C locale) that it contains...
> # ^          at BOL...
> # [0-9]   require a digit
> # *          ... and digits following
> # $          ...until we reach EOL
> # ... but...:
> echo "$x" | grep -c '^[0-9]*$'
> 1             <== WHY?
> it works if i remove the *, but this would only work for 1digit nrs.
> is this a grep bug or my false understanding of how '*' works?
>
you want to use + (1 or more) instead of * (0 or more) and for that 
youll need to add -E mode to grep as well