Re^2: hash question

Thanks all for your prompt responses. I don't understand why the address has been changed between "Sa:" and "Sb:" by assigning another anonymous hash. I expected "Sb:" would be the same as "Sa:", since i was using the same var. I changed the code a bit:

sub do_something
{
    my $h = shift;

    print '1. do_something(): ' . $h . "\n";

    %{ $h } = ( a => 'alpha', b => 'beta' );

    print '2. do_something(): ' . $h . "\n";
    print Dumper( $h );
}
[download]

and it shows:

hash = HASH(0x490d30)
1. do_something(): HASH(0x490d30)
2. do_something(): HASH(0x490d30)
{
  'a' => 'alpha',
  'b' => 'beta'
}
{
  'a' => 'alpha',
  'b' => 'beta'
}
[download]

The address is not changed ... the behaviour is not consistent to me :)

Comment on Re^2: hash question Select or Download Code

Replies are listed 'Best First'.
Re^3: hash question by andal (Hermit) on May 27, 2016 at 07:44 UTC
You are missing one slight catch. When you do `my $hash = shift;` [download] you get pointer to the object passed to the function. When you do `$hash = {a => "alpha"};` [download] you create new object and store its pointer in the variable that previously contained pointer to another object. Finally, when you do `%{ $hash } = ( a => 'alpha', b => 'beta' );` [download] then you take object pointed by the variable and store in that object new set of keys. You have to distinguish "objects" and "pointers to objects", the latter are called "references" in perl, then things may become more consistent for you.	[reply] [d/l] [select]
Re^3: hash question by soonix (Canon) on May 27, 2016 at 07:00 UTC
You are printing the content of the variable. It's two different variables. If you `print` the variables directly instead of via Dumper, `my $hash = {}; do_something( $hash ); print Dumper( $hash ); sub do_something { my $hash = shift; $hash = { a => 'alpha', b => 'beta', }; print Dumper( $hash ); }` [download] you'll get two different addresses: `HASH(0x3c4a4)HASH(0x3c39c)` [download] Update: actually, this is also only the contents. Best answer so far to your OP is this by AnomalousMonk	[reply] [d/l] [select]
Re^3: hash question by AnomalousMonk (Archbishop) on May 27, 2016 at 16:12 UTC
I don't understand why the address has been changed between "Sa:" and "Sb:" by assigning another anonymous hash. I expected "Sb:" would be the same as "Sa:", since i was using the same var. [Emphasis added] That's like saying In the code `$x = 4;` `$x = 7;` I don't understand why the value of `$x` is 7 after the second assignment to `$x`: I'm using the same variable. As others have noted, it's vital to distinguish between a variable, the content of the variable, and, if the content of the variable is a reference, the referent of the reference. References are tricky. Give a man a fish: `<%-{-{-{-<`	[reply] [d/l] [select]


Think about Loose Coupling
	PerlMonks