flip-flop interpolation

[Anonymous Monk]

print if 2 .. 4 # works
print if $i .. $j # doesn't work
# how to make second to work?

[choroba]

Range Operators says:

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, if you're not using constants, you have to do the comparison yourself:

print if ($. == $i) .. ($. == $j); # The parentheses are optional.

[Anonymous Monk]

Thank you.
And can't I say to Perl to use my $i and $j as constants?

[SuicideJunkie]

That is like asking to draw a red line using green ink :)

The way to do it is to use the green ink to define a universe containing hardcoded red lines:

#untested, and quite silly
eval "print $_ if $i .. $j;";

[johngg]

You could use the constant pragma but it is not really saving you anything.

$ seq 1 6 > lines
$ cat lines
1
2
3
4
5
6
$ perl -ne '
> use constant { I => 3, J => 5 };
> print if I .. J;' lines
3
4
5
$

I hope this is of interest.

[Anonymous Monk]

And can't I say to Perl to use my $i and $j as constants?

You can say anything you want :) but scalar literals are not scalar variables

[Anonymous Monk]

how to make second to work?

Don't try to make it work, use something else :)

saving matching pattern with flip-flop operator, Flipin good, or a total flop?, Why doesn't the flip-flop operator work in all scalar contexts?, The Scalar Range Operator, grep flip flop (range operator)

[GotToBTru]

$i = 2;
$j = 4;
print "$_\n" for ($i..$j);

Output:

2
3
4

Update: I missed what was obvious to choroba. Not so much a flip-flop as a filter:

# show lines 2 thru 4
while (<>) {
  print if 2..4
}

# this will print every line
$i = 2; 
$j = 4;
while (<>) {
  print if $i..$j
}