Perl: Why can I call subroutine A recursively, but I can't call A from B, which is called from an earlier A? - perl

Question

Try to "Play again" by answering Y when asked. Or am I just missing something completely obvious?
#!/usr/bin/perl
sub calc() {
$circ = 2 * $radius * 3.141592654;
print "The circumference is $circ\nPlay again (Y/n)? ";
$again = <STDIN>;
if ($again eq "Y") {
dialog();
} else {print "Bye!\n";
}
}
sub dialog {
print "What's the radius of the circle you'd like the circumference of?\n> ";
$radius = <STDIN>;
if ($radius eq "\n") {
print "That was just a blank line!\n";
dialog();
} elsif ($radius < 0) {
print "That was negative! Let's assume radius is 0.\n";
$radius = 0;
calc();
} else {calc();
}
}
dialog();

Answer 1

You don't chomp your input from STDIN
chomp($again = <STDIN>);
Also, always include use strict; and use warnings; at the top of each and EVERY perl script.
Finally, instead of doing recursion, a simple loop is sufficient. The following is a cleaned up version of your code:
#!/usr/bin/perl
use strict;
use warnings;
while (1) {
print "What's the radius of the circle you'd like the circumference of?\n> ";
chomp(my $radius = <STDIN>);
if ($radius eq "") {
print "That was just a blank line!\n";
next;
}
if ($radius < 0) {
print "That was negative! Let's assume radius is 0.\n";
$radius = 0;
}
my $circ = 2 * $radius * 3.141592654;
print "The circumference is $circ\nPlay again (Y/n)? ";
chomp(my $again = <STDIN>);
if ($again ne "Y") {
print "Bye!\n";
last;
}
}