Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

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

Replies are listed 'Best First'.
Re: flip-flop interpolation
by choroba (Archbishop) on May 13, 2015 at 10:02 UTC
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.
لսႽ† ᥲᥒ⚪⟊Ⴙᘓᖇ Ꮅᘓᖇ⎱ Ⴙᥲ𝇋ƙᘓᖇ
Thank you.
And can't I say to Perl to use my \$i and \$j as constants?

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;";

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.

Cheers,

JohnGG

Re: flip-flop interpolation
by Anonymous Monk on May 13, 2015 at 10:25 UTC

how to make second to work?

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

Re: flip-flop interpolation
by GotToBTru (Prior) on May 13, 2015 at 18:25 UTC
```\$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
}