Re: Escaping grep

On Sun, 21 Feb 2021 15:35:55 -0600, "Michael F. Stemper"
<[email protected]> wrote:

> I wanted to see which LaTeX packages I was using for a particular
> project, so I did:
> 
> user@host$ grep -h "\\usep" *tex | sort | uniq
> % \usepackage{array} % for math mode columns in tabular
> % \usepackage{graphicx}
> % \usepackage{longtable}
> % \usepackage{mathrsfs} % for mathscr
> % \usepackage{tikz-cd} % for commutative diagrams
> \usepackage{amsmath} % for implication and align*
> \usepackage{amssymb} % for black-board set names
> \usepackage{fancyhdr}
> \usepackage{geometry}
> \usepackage{parskip}
> user@host$
> 
> However, that showed unused packages as well. So I tried
> anchoring my grep to the beginning of the line, and got:
> 
> user@host$ grep -h "^\\usep" *tex | sort | uniq
> user@host$
> 
> Why isn't this working? What am I overlooking? How
> should I have said it?

There is no issue with the grep expression itself. The problem is
with bash handling of strings: You use " (double quotes) around
your expression, asking bash to interpret the contents.

bash will replace "^\\usep" with "^\usep" *before passing it to grep*.
grep then will remove special meaning for following char (u) - no
special meaning in this case.
The result is that grep will just look for lines starting with "u".

You have 2 solutions:
1) use quotes instead of double quotes: '^\\usep', so that bash does
not change anything.
2) using double quotes, force bash to preserve 2 backslash with the
string "^\\\\usep"

Sorry for late answer, I just joined this group.

And yes, it is a pure 'bash' question, not a 'grep' question as I saw in
another answer :)

bodi.