Print how many substitutions took place in a Perl

Question

I've used web search, found similarly titled question How many substitutions took place in a Perl s///g? and tried to use it to print the number but have not been able to succeed.

My initial code was

perl -0777 -i.original -pe 's-\r\n-\n-igs' test.txt

When I tried

perl -0777 -i.original -pe "$c=s-\r\n-\n-igs;say qq'$c'" test.txt

I got nothing - no output and no replacements, when I tried

perl -0777 -i.original -pe '$c=s-\r\n-\n-igs;print qq($c\n)' test.txt

(print similar to other one-liners I used before) I got empty string in standard output but 454847 added as the beginning of file (and proper replacements).

I understand =~ is not needed in my case (What does =~ do in Perl?), so what is wrong with my code? How to print number of replacements made?

Answer 1

Because of -i, the default output handle isn't STDOUT but the output file. To print to STDOUT, you'll need to do so explicitly.

If using the cmd shell,

perl -0777pe"CORE::say STDOUT s/\r//g" -i.original test.txt

If using sh or similar,

perl -0777pe'CORE::say STDOUT s/\r//g' -i.original test.txt

Notes:

• s/// is more idiomatic than s---
• /i is useless here.
• /s is useless here.
• There no point in checking for Line Feeds. s/\r//g will work just as fine as s/\r\n/\n/g.
• There's point in using a variable for the count, especially if using say instead of print.
• One must use CORE::say instead of say for backwards compatibility reasons unless use feature qw( say ); or equivalent is used.
• Neither s/\r\n/\n/g nor s/\r//g will work with a Windows build of Perl, and there's no way to do what you want when using -i on a Windows build of Perl. However, you are using a unix build of Perl (since MSYS is a unix emulation environment), so it's not an issue.