How to grab last character in a regex with grep

Question

I'm trying to grab the last space and what follows it on a line using grep.

This grab me the first space :

echo "toto tata titi" | grep -o " .*$"

In Java I would have used the non-greedy operator but it does not seem to work :

echo "toto tata titi" | grep -o " .*?$"

It return nothing

The expected result is titi.

Answer 1

Replace . with [^ ], which matches everything but space. Then it can be greedy.

echo "toto tata titi" | grep -o " [^ ]*$"

(If you want grep to use extended regexes, either use egrep or grep -E.)