I want to replace $foo with bar. This works fine.
$ echo '$foo' | sed 's/$foo/bar/' bar
But this command does not work fine when I use the -r option.
$ echo '$foo' | sed -r 's/$foo/bar/' $foo
Why doesn’t this work.
Here is an example of what works with -r option.
$ echo '$foo' | sed -r 's/$foo/bar/' bar
The real question is: Why does $ need to be escaped only while using the -r option. What would $ mean otherwise with the -r option?
I am using a Debian Linux system.
Advertisement
Answer
In extended regular expressions (ERE), that is, under -r, $ means the end of the line:
$ echo '$foo' | sed -r 's/foo$/bar/' $bar
If you want it to mean something else, it has to be escaped:
$ echo '$foo' | sed -r 's/[$]foo$/bar/' bar
Documentation
man 7 regex explains that, in Extended Regular Expressions (ERE), the $ matches at the end of the line:
‘$’ (matching the null string at the end of a line)
The same man page goes on to explain that in Basic Regular Expressions (BRE), which is what you get without -r, its meaning is more complicated:
Obsolete (“basic”) regular expressions differ in several respects. ‘|’, ‘+’, and ‘?’ are ordinary characters and there is no equivalent for their functionality. The delimiters for bounds are “{” and “}”, with ‘{‘ and ‘}’ by themselves ordinary characters. The parentheses for nested subexpressions are “(” and “)”, with ‘(‘ and ‘)’ by themselves ordinary characters. ‘^’ is an ordinary character except at the beginning of the RE or(!) the beginning of a parenthesized subexpression, ‘$’ is an ordinary character except at the end of the RE or(!) the end of a parenthesized subexpression, and ‘*’ is an ordinary character if it appears at the beginning of the RE or the beginning of a parenthesized subexpression (after a possible leading ‘^’).