I have a subroutine that takes the value of $x and gives that value of $a. I can't access the value of $a outside the subroutine, though, since it tells me $a is undefined. I only learned about subroutines yesterday, so I assume there's something about them I'm missing.
sub rout {
if(@_ == 1) {
my $a = 3;
} else {
my $a = 5;
}
}
my $x = 1;
rout($x);
print $a;
First of all, don't use $a and $b. They're a bit special because of their use by sort.
Secondly, @_ == 1 checks the number of arguments provided by the caller. This probably not what you wanted to check. You probably wanted to check the value of the first argument: $_[0] == 1.
The issue is that you are creating a new lexically-scoped variable, assign a value to it, then immediately leave the scope. Your variable is destroyed as soon as you create it! Declare a single variable in the outermost scope where it's needed.
my $y;
sub rout {
my ($x) = @_;
if ($x == 1) {
$y = 3;
} else {
$y = 5;
}
}
my $x = 1;
rout($x);
print "$y\n";
That said, returning a value would make more sense here.
sub rout {
my ($x) = @_;
if ($x == 1) {
return 3;
} else {
return 5;
}
}
my $x = 1;
my $y = rout($x);
print "$y\n";
my $a will not trigger use strict to report a variable out of scope. A subtle error that may stump new users.