Bash Special Characters – Can’t Use Exclamation Mark (!) in Bash

The ! character invokes bash's history substitution (enabled by default in interactive shells). When followed by a string (as in your failing example) it tries to expand to the last history event that began with that string. Just like $var gets expanded to the value of that string, !echo would expand to the last echo command in your history.

Space is a breaking character in such expansions. First note how this would work with variables:

# var="like"
# echo "$var"
like
# echo "$"
$
# echo "Do you $var frogs?"
Do you like frogs?       <- as expected, variable name broken at space
# echo "Do you $varfrogs?"
Do you?                  <- $varfrogs not defined, replaced with blank
# echo "Do you $ var frogs?"
Do you $ var frogs?      <- $ not a valid variable name, ignored

The same thing will happen for history expansion. The bang character (!) starts off a history replacement sequence, but only if followed by a string. Following it with a space make it literal bang instead of part of a replace sequence.

You can avoid this kind of replacement for both variable and history expantion by using single quotes. Your first examples used single quotes and so ran fine. Your last examples are in double quotes and thus bash scanned them for expantion sequences before it did anything else. The only reason the first one didn't trip is that that the space is a break character as shown above.

The rules are different for single quotes versus double quotes.

For the reason you show, double quotes can't be used reliably in bash, because there's no sane way to escape an exclamation mark.

$ grep -oP "\\(.*(?!word).*right"
bash: !word: event not found

$ grep -oP "\\(.*(?\!word).*right"
grep: unrecognized character after (? or (?-

The second is because bash passes through \! rather than ! to grep. Showing this:

$ printf '%s' "\!"
\!

When you tried single quotes, the double backslash doesn't mean an escaped backslash, it means two backslashes.

$ printf '%s' '\\(.*(?!word).*right'
\\(.*(?!word).*right

Inside single quotes, everything is literal, and there are no escapes, so the way to write the regular expression you're trying is:

$ grep -oP '\(.*(?!word).*right'