Hello glendeni,

Yes, /\s/ does match a newline character:

13:41 >perl -wE "say qq[Matched >$1<] if qq[abc2345] =~ /([0-9]+)\s/;

13:41 >perl -wE "say qq[Matched >$1<] if qq[abc2345 ] =~ /([0-9]+)\s/;
Matched >2345<

13:42 >perl -wE "say qq[Matched >$1<] if qq[abc2345\n] =~ /([0-9]+)\s/
+;
Matched >2345<

13:42 >perl -v

This is perl 5, version 26, subversion 0 (v5.26.0) built for MSWin32-x
+64-multi-thread-ld
[download]

For the meaning of \s in a regex, see “Whitespace” in perlrecharclass#Backslash-sequences.

Hope that helps,

Thanks for confirming my understanding. Looking for something else to explain my result, realized I had fixated on regexp but actual problem was input of the lines from a file followed by chomp to strip the end-of-lines.

As Athanasius's examples here demonstrate, the  \s class requires a whitespace character (if  * ? quantifiers are not present), and that's not what you have if you chomp off the newline. You may be interested to investigate the  $ \z \Z line end anchors. In your particular case,  \Z (big-Z) might be useful to "Match only at end of string, or before newline at the end" (see perlre); a  \Z would match match the end of either a chomp-ed or un-chomp-ed line (assuming you're chomp-ing newlines).

Update: Fixed  ^ vice  $ in "... the  ^ \z \Z line end anchors." per Lotus1's /msg. Thanks!

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