I'm studying bash programming , in particular the regex and I found this code:
numpat='^[+-]([0-9]+)$'
strpat='^([a-z]*)\1$'
read stringa
if [[ $stringa =~ $numpat ]]
then
echo "numero"
echo numero > output
exit ${BASH_REMATCH[1]}
elif [[ $stringa =~ $strpat ]]
then
echo "echo"
echo echo > output
exit 11
fi
and I don't understand what means \1 in this line:
strpat='^([a-z]*)\1$'
\1 is a backreference. It matches whatever was matched by the first capture group ([a-z]*).
So the pattern ^([a-z]*)\1$ matches a string that built from a substring that's repeated twice, such as foofoo. The capture group matches the first foo, and the backreference matches the second foo. But if the string is foobar, the backreference never matches anything, because it can't find another repetition of any of the initial strings.
You can allow any number of repetitions by using the + quantifier after \1. This matches it one or more times.
* as a pattern makes it greedy so it will probably not match anything else. If I test it, it doesn't work: [[ "foofoo" =~ ^([a-z]*)\1$ ]] && echo "yes" || echo "no" returns no.
^([a-z]*)\1{4}$
1.On cygwin, which uses newlib, \1 matches only 1.
if [[ a1 =~ $strpat ]]; then echo YES; fi # YES
foofoo but not foobar. That's weird! I have GNU bash, version 4.3.42(1)-release.
man 3 regex, and back reference is not in the ERE standard, although it may be supported in some implementations.grep back references work just fine on my machine.