How many substitutions took place in a Perl s///g?

regex perl substitution
musiKk picture musiKk · Jun 23, 2010 · Viewed 8.1k times · Source

Small example:

perl -e '$s="aaabbcc";$c=()=$s=~/a/g;print"$c\n$s\n"' (m//g) outputs

3
aaabbcc

whereas perl -e '$s="aaabbcc";$c=()=$s=~s/a/x/g;print"$c\n$s\n"' (s///g) outputs

1
xxxbbcc

I'd like to do both things at once without having to match first: substitute and know the number of substitutions. Obviously a s///g does not return the number of substitutions in scalar context--unlike m//g does with matches. Is this possible? If yes, how?

perlre, perlvar and perlop provided no help (or I just couldn't find it).

Answer

Michael Carman picture Michael Carman · Jun 23, 2010

s/// does return the number of substitutions made in scalar context. From perlop (emphasis added):

s/PATTERN/REPLACEMENT/msixpogce
Searches a string for a pattern, and if found, replaces that pattern with the replacement text and returns the number of substitutions made. Otherwise it returns false (specifically, the empty string).

Your problem is that you didn't call s/// in scalar context. You called it in list context and then evaluated the assignment (to an empty list) in scalar context. A list assignment in scalar context returns the number of elements produced by the right-hand side of the expression. Since s/// returns a single value (in both list and scalar context) the number of elements is always one even if the s/// didn't do anything.

perl -E "$s='aaabbcc'; $c=()=$s=~s/x/y/g; say qq'$c-$s'"  # prints "1-aaabbcc"

To call s/// in scalar context, omit the =()= pseudo-operator.

perl -E "$s='aaabbcc'; $c=$s=~s/a/x/g; say qq'$c-$s'"  # prints "3-xxxbbcc"

Related questions

How to use a variable in the replacement side of the Perl substitution operator?

I would like to do the following: $find="start (.*) end"; $replace="foo \1 bar"; $var = "start middle end"; $var =~ s/$find/$replace/; I would expect $var to contain "foo middle bar", but it does not work. Neither does: $replace='foo \1 bar'; …

Read More
How do I perform multiple replacements with Perl?

I have Perl code: my $s = "The+quick+brown+fox+jumps+over+the+lazy+dog+that+is+my+dog"; I want to replace every + with space and dog with cat. I have this regular expression: $s =~ s/\+(.*)dog/ ${1}cat/g; …

Read More
Inline regex replacement in perl

Is there a way to replace text with a regex inline, rather than taking the text from a variable and storing it in a variable? I'm a perl beginner. I often find myself writing my $foo = $bar; $foo =~ s/regex/…

Read More