4

I have a file with the following 3 lines:

file.txt

aaa|
bbb|
ccc|

and another file with one line:

regex.txt

^aaa\|

$grep ^aaa\| file.txt yields:

aaa|

$grep -f regex.txt file.txt yields:

aaa|
bbb|
ccc|

Why are the results different for grep -f and grep?

$grep -V
grep (GNU grep) 3.1

$lsb_release -a
No LSB modules are available.
Distributor ID: Ubuntu
Description:    Ubuntu 18.04.1 LTS
Release:    18.04
Codename:   bionic

2 Answers 2

6

You're using grep with basic regular expressions. grep ^aaa\| file.txt is the same as typing grep "^aaa|" file.txt. While reading the file grep -f regex.txt file.txt is the same as grep "^aaa\|" file.txt., the escaped | means match ^aaa or "" which matches anything.

1
  • 2
    In particular, on the command line the shell processes and removes the backslash (or the double-quotes you propose). While if the pattern is in a file, there is no shell involved, so greps regex engine sees the backslash. It's probably easiest to single-quote any such patterns on the shell command line, to ensure minimal processing by the shell. Also, it should be noted that \| for alternation in BRE is a GNU thing, and not standard.
    – ilkkachu
    Jul 23, 2019 at 6:23
0

When using regexp in file, you do not need the \ (escape). When in file it is treated as the \ char. In command it is treated as escape and ignored (due to the lack of quoting of the expression).

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