escaping 如何为 sed 工作

Question

The following sed command

echo '.' | sed "s/\\./foo/"

substitutes . with foo , as expected. However, if we escape the non-alphanumeric . in the above command

echo '.' | sed "s/\\\./foo/"

prints barely . , whereas foo is expected. sed should match the character . literally, but it doesn't. I cannot understand what is happening with the dot. I believe that I should simply put a backslash in front of every non-alphanumeric character in bash, if a string is double-quoted. A dot is a non-alphanumeric character so what is wrong about escaping it and why does it produce a different result?

Answer 1

This is because how backslashes work in bash double quote " escaping.

echo "\\"
\
echo "\."
\.
echo "s/\\./foo/"
s/\./foo/
echo "s/\\\./foo/"
s/\\./foo/

From man bash :

Within double quotes, the backslash retains its special meaning only when
followed by one of the following characters: $, `, ", \,or <newline>.

So in the first case, sed gets s/\./foo/ and interprets it as "replace a dot with foo". In the second case sed gets s/\\./foo/ and interprets it as "replace a backslash and one other character with foo.

You better use single quote escaping in this case:

echo 's/./foo/'
s/./foo/
echo 's/\./foo/'
s/\./foo/

which is probably what you wanted.

Answer 2

In your second one, the . is still a literal dot, but the regular expression only replaces the two-character sequence \. (not . by itself) with foo :

$ echo '=\.=' | sed "s/\\\./foo/"
=foo=