Perl Flip Flop operator and line numbers

Question

I noticed this while looking at another question...

If I have a script like this:

while (<>) {
  print if 5 .. undef;
}

It skips lines 1..4 then prints the rest of the file. However if I try this:

my $start_line = 5;
while (<>) {
  print if $start_line .. undef;
}

It prints from line 1. Can anyone explain why?

Actually I'm not even sure why the first one works.

Hmmm looking further into this I found that this works:

my $start = 5;
while (<>) {
  print if $. ==  $start .. undef;
}

So the first version magically uses $. which is the line number. But I don't know why it fails with a variable.

score 10 · Accepted Answer · answered Jun 08 '14 at 11:37

10

The use of a bare number in the flip-flop is treated as a test against the line count variable, $.. From perldoc perlop:

If either operand of scalar ".." is a constant expression, that operand is considered true if it is equal (==) to the current input line number (the $. variable).

So

print if 5 .. undef;

is "shorthand" for:

print if $. == 5 .. undef;

The same is not true for a scalar variable as it is not a constant expression. This is why it is not tested against $..

answered Jun 08 '14 at 11:37

Zaid

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2

Perl is sometimes more like a 'magic spell book' than a language. – John C Jun 08 '14 at 11:55
2

@JohnC : That is even truer for the English language :) – Zaid Jun 08 '14 at 12:13

Perl Flip Flop operator and line numbers

1 Answers1