Passing a scalar reference in Perl

Question

I know that passing a scalar to a sub is actually passing the reference, but since I am new to perl I still did the following test:

#!/usr/bin/perl
$i = 2;
subr(\$i);
sub subr{
    print $_[0]."\n";
    print $$_[0]."\n";
}

I thought the first line is going to print an address and the second line is going to give be back the number, but the second one is a blank line. I was pointed by someone one else to do this: ${$_[0]} and it prints the number. But she didn't know the reason why without {} it is not working and why it is working with {}. So what has happened?

Answer 1

It's because your second print statement is equivalent to doing this...

my $x = $$_; print $x[0];

When what you want is

my $x = $_[0]; print $$x;

In other words, the de-referencing occurs before the array subscript is evaluated.

When you add those curl-wurlies, it tells perl how to interpret the expression as you want it; it will evaluate $_[0] first, and then de-reference to get the value.

Answer 2

It's an order-of-evaluation thing.

  $$_[0] is evaluated as {$$_}[0]

This is the 0th element of the reference of the scalar variable $_. It's taking the reference first and then trying to find the 0th element of it.

  ${$_[0]}

This is a reference to the 0th element of the array @_. It's finding the 0th element first then taking a reference of that.

If you set use strict and use warnings at the top of your code you'll see plenty of warnings about undefined values from your first attempt.