How do you disable the special meaning of the ^ character in grep [] syntax?

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Suppose, I have a file like below:

aaa1bbb
aaa^bbb
aaa\bbb
aaa%bbb
aaa*ccc
aaa(ccc
aaa)bbb

I want to find all lines that contain either ^ or % and then bbb. My expected results:

aaa^bbb
aaa%bbb

When I execute grep '[^%]bbb' t1 I have received

aaa1bbb
aaa^bbb
aaa\bbb
aaa)bbb

which is completely correct. ^ is metacharter in grep []syntax. I have all the results except %bbb regex. I understand this. I have tried to disable metacharacter with \ like this: grep '[\^%]bbb' t1 I have received:

aaa^bbb
aaa\bbb
aaa%bbb

Why I got line aaa\bbb in the results? The problem is: How do you disable the special meaning of the ^ character in grep [] syntax? Is it possible without using fgrep or other workarounds?

grep regex
1 Answers

To make a ^ literal char inside a bracket expression, put it NOT at the start of it:

grep '[%^]bbb' t1

This is part of the so-called "smart placement" technique, e.g. to use a ] inside a bracket expression it must be placed at the start of a bracket expression, and - should be either at the start or end of the bracket expression.

See an online demo:

#!/bin/bash
s='aaa1bbb
aaa^bbb
aaa\bbb
aaa%bbb
aaa*ccc
aaa(ccc
aaa)bbb'
grep '[%^]bbb' <<< "$s"

Output:

aaa^bbb
aaa%bbb
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