in reply to matching lines in a long string

Hello Cristoforo,

Within the look-ahead assertion (?=$), the metacharacter $ means:

Match the end of the string (or before newline at the end of the string; or before any newline if /m is used) (from “Metacharacters” in perlre#The-Basics)

Since it comes after the newline within the regex pattern — and your string contains no consecutive newlines — it can match only at the very end of the string. This explains the behaviour you’re seeing.

If you remove the look-ahead:

for my $line (split/(?m)\n/, $s) {

the output is displayed line-by-line, as expected.

Hope that helps,

Athanasius <°(((><contra mundum Iustus alius egestas vitae, eros Piratica,

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Re^2: matching lines in a long string
by AnomalousMonk (Archbishop) on May 30, 2020 at 14:52 UTC

    I disagree slightly with your narrative. A split pattern of  /(?m)\n(?=$)/ will split on any newline that is followed immediately by a newline (i.e., consequtive newlines) or on a newline at the end of the string (producing an empty string). This can be seen to be happening with modified text (and with a split LIMIT of -1 to preserve trailing null fields):

    c:\@Work\Perl\monks>perl use strict; use warnings; use feature 'say'; my $s = <<'EOF'; int t; //variable t t->a=0; //t->a does;; something printf("\nEnter the Employee Name : "); scanf("%s", ptr->name); return 0; EOF for my $line (split/(?m)\n(?=$)/, $s, -1) { say '<' . $line . '>', "\n"; } __END__ <int t; //variable t t->a=0; //t->a does;; something> < printf("\nEnter the Employee Name : "); scanf("%s", ptr->name); return 0;> <>

    I agree that the simplest and best approach is to split on what the OPer wants to split on, i.e., newlines, and forget about the  /m modifier and everything else.


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