Re^2: Taking reference to function

[Freezy]

This behavior confuses me. EG:

my $string = "Hello.";
my $string_ref = \$string;

$string = "Bacon!";
print $$string_ref;

The above prints "Bacon!", as my intuition expects. Why does the coderef point to the old block?

[ELISHEVA]

Because the symbol foo and the coderef are different things. *foo = ... assigns a new coderef to the symbol foo. It doesn't redefine the coderef for the original sub foo, which is kind of like a constant.

\&foo takes the address of a constant, not a variable. So it is analogous to \"Hello World" rather than \$some_var_holding_a_string. You can't reassign it any more than you can reassign the address of the string literal "Hello World".

Well, almost so. Sub "literals" are more complex than a string. Redefine a string and you get a different string. Redefine a sub and you get a different association between sub name and body. So if you *really*, *really* were set on redefining sub foo itself, you could, but (a) you have to nudge Perl into compiling code at runtime and (b) you will get a warning about redefining things that are not supposed to be redefined:

use strict;
use warnings;

sub sayHello { print "Hello, world!\n"; }
sayHello();
eval q{sub sayHello { print "Bonjour, le monde!\n"; }};
sayHello();

which outputs

Hello, world!
Subroutine sayHello redefined at (eval 1) line 1.
Bonjour, le monde!

Best, beth

Update: added some explanation about difference between sub and string literals.

[syphilis]

Why does the coderef point to the old block?

That's not a code reference, it's a scalar reference.
$string_ref is a reference to the scalar, $string. Doing a print $$string_ref; will print out whatever's *presently* in $string.

Update: Oops ... looking at Freezy's post in the context of the entire thread, I think I might've missed the point somewhat in my reply.

Cheers,
Rob