Use grep to search for words beginning with non-word characters

Question

The contents of the file testing.txt are:

ls -a
cmake --verbose
verbose

I want to use grep to look through this file and find only the word beginning with "--" i.e. the word "--verbose"

However using the following patterns as an argument for grep does not work:

$ cat testing.txt | grep -- 
Usage: grep [OPTION]... PATTERN
   [FILE]... Try 'grep --help' for more information.

$ cat testing.txt | grep -
ls -a
cmake --verbose

$ cat testing.txt | grep '--v'
grep (GNU grep) 3.1
Copyright (C) 2017 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>.
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.

Written by Mike Haertel and others, see <http://git.sv.gnu.org/cgit/grep.git/tree/AUTHORS>.

$ cat testing.txt | grep ver
cmake --verbose
verbose

$ cat testing.txt | grep '-ver'
ls -a

• grep thinks that all arguments beginning with a -- are options? How do you prevent this so that grep can search for a pattern (in a file) that begins with "--"?
• the last attempt uses the pattern "-ver" so that grep does not think the pattern is an option, but then grep does not match the word "--verbose" in the file even though it contains the pattern "-ver". What causes this behavior?

Answer 1

The string -- is special for most utilities when it occurs on the command line. It signals the end of options to the command line argument parser. It is used in situations where you may want to pass a filename that starts with a dash, as in rm -- -f (to delete a file called -f in the current directory).

To use -- as a pattern with grep, tell the utility explicitly that it is a pattern:

grep -e --

The -e option to grep takes an option argument which is the pattern that you want grep to search with.

You could also use

grep -- --

Here, grep knows that the second -- is the pattern, because the first -- says it can't be an option.

---

Your last pipeline returns ls -a because that's a line in the file that does not include an r. The command grep -ver may also be written grep -v -e r, i.e., "extract all the lines that do not (-v) match r (-e r)".

Answer 2

To answer the question in the title which I interpret as meaning: report the whitespace-delimited words that do not start with an alphanumeric or underscore (the \w in some regexp engines), you could do with GNU grep provided it's built with PCRE support:

grep -Po '(?<!\S)[^\w\s]\S*'

That is a character other (^) than a word one (\w) or whitespace one (\s) provided it's not preceded ((?<!...)) by a non-whitespace (\S), and followed by any number (*) of non-whitespace (\S).

Which on your input returns:

-a
--verbose

Answer 3

With GNU grep you an approach this problem by means of positive lookarounds as shown:

 $ grep -oP '(?:(?!\h)\W)+\w+' inp

To be read as, from where i stand in the string, i see a nonword follow a word. But, that nonword precludes any whitespace.

Output:

 -a
 --verbose