No, that's completely wrong. chomp removes what's in $/ from the end of its arguments. $/ is equal to \n (by default) in Windows, so chomp only removes \n (by default) on Windows too. 

die("\\n ne LF\n")
   if "abc\n" ne "abc\x0A";

print("\\n eq LF\n");

$s = "abc";
$l1 = length($s);
chomp($s);
$l2 = length($s);
if ($l1 - $l2 == 0) {
   print("abc unchanged\n");
} else {
   print("???\n");
}

$s = "abc\n";
$l1 = length($s);
chomp($s);
$l2 = length($s);
if ($l1 - $l2 == 1) {
   print("\\n removed from abc\\n\n");
} elsif ($l1 - $l2 == 0) {
   print("abc\\n unchanged\n");
} else {
   print("???\n");
}

$s = "abc\r\n";
$l1 = length($s);
chomp($s);
$l2 = length($s);
if ($l1 - $l2 == 2) {
   print("\\r\\n removed from abc\\r\\n\n");
} elsif ($l1 - $l2 == 1) {
   print("\\n removed from abc\\r\\n (\\r kept)\n");
} elsif ($l1 - $l2 == 0) {
   print("abc\\r\\n unchanged\n");
} else {
   print("???\n");
}

__END__
ActivePerl v5.8.0 on WinXP
--------------------------
\n eq LF
abc unchanged
\n removed from abc\n
\n removed from abc\r\n (\r kept)

[download]

Perl translates CR+LF to LF for you when you read from a file in Windows (unless you use binmode on that handle). Similarly, Perl translates LF to CR+LF when writting to a file handle in Windows (unless you use binmode on that handle).

The reason to use chomp is to handle the case where a line without the line terminator is read in. chop would fail to perform as desired there, but chomp wouldn't. That has nothing to do with CR+LF vs LF.

Update: Added underlined text to clarify my intended message, at GrandFather's recommendation.

In reply to Re^2: Problem with <STDIN> by ikegami
in thread Problem with <STDIN> by sivaramanm

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