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Re^2: Taking reference to function

by Freezy (Scribe)
on Apr 16, 2009 at 06:01 UTC ( #757848=note: print w/replies, xml ) Need Help??

in reply to Re: Taking reference to function
in thread Taking reference to function

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?

Replies are listed 'Best First'.
Re^3: Taking reference to function
by ELISHEVA (Prior) on Apr 16, 2009 at 06:24 UTC

    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.

Re^3: Taking reference to function
by syphilis (Archbishop) on Apr 16, 2009 at 06:20 UTC
    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.


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