What is the difference between \1 and $1 in a Perl regex?

Question

What is the difference of doing \\1 as opposed to $1 if any, or are they interchangeable in all situations.

Example:

s/([a-z]+),afklol/$1,bck/;
#against
s/([a-z]+),afklol/\1,bck/;

They both give the same result but is there any difference?

Answer 1

Straight from perldoc perlre :

Warning on \\1 vs $1

Some people get too used to writing things like:

 $pattern =~ s/(\\W)/\\\\\\1/g; 

This is grandfathered for the RHS of a substitute to avoid shocking the sed addicts, but it's a dirty habit to get into. That's because in PerlThink, the righthand side of an "s///" is a double- quoted string. "\\1" in the usual double-quoted string means a control-A. The customary Unix meaning of "\\1" is kludged in for "s///". However, if you get into the habit of doing that, you get yourself into trouble if you then add an "/e" modifier.

 s/(\\d+)/ \\1 + 1 /eg; # causes warning under -w 

Or if you try to do

 s/(\\d+)/\\1000/; 

You can't disambiguate that by saying "{1}000", whereas you can fix it with "${1}000". The operation of interpolation should not be confused with the operation of matching a backreference.

Certainly they mean two different things on the left side of the "s///".

Answer 2

I only use \\1 on the rare occasions where I need the match to include a repeat of an earlier part of the pattern. eg

/(foo|bar)baz\1/

This would match "foobazfoo" or "barbazbar" but not "foobazbar". You can't use $1 there.

In all other circumstances I use $1 and kin, because there are too many gotchas to using the other notation. See perlre for all the gory details.

Answer 3

They give the same result, but $1 retains its value while \\1 does not. Try this:

s/([a-z]+),afklol/$1,bck/;
print "\$1 is $1\n";
#against
s/([a-z]+),afklol/\1,bck/;
print "\\1 is \1\n";