I can't store it somewhere. I have a routine to which a user can pass a coderef. And in that routine i want to know the full name of that coderef.
-Sharad
But in that case isn't it just a design issue, to me it would make more sense to pass the name of the routine and call it directly from the symbol table rather than passing the code ref, or pass both the ref and its' name, since they are really independent pieces of information you don't *need* one to use the other, it would limit the ability of the sub if you are passing a code ref, and it is expecting a code ref & and sub name in the same argument, since presumably if you want the code to call it as a subroutine an anonymous code ref should work in the same manner as any other code ref, but it won't have a name, so you are really wanting "a code ref with a name" as your argument, but at that point it makes more sense to break that into "a code ref, and a name"....
But then, obviously I haven't seen the rest of the code......
http://danconia.org
-----Original Message----- From: Wiggins d'Anconia [mailto:[EMAIL PROTECTED] Sent: Wednesday, October 22, 2003 6:23 PM To: Gupta, Sharad Cc: [EMAIL PROTECTED] Subject: Re: FQN from references
Gupta, Sharad wrote:
Hi All,
How do i know the full name of a variable from its reference??.
Say i have a routine defined like this:
package Bar; sub foo {};
And then i say: my $x = \&foo;
Now from this coderef ($x) i want to know its fully
qualified name which would be somthing like Bar::Foo.
Is there a way to do that.??.
Where do you want to know it? Presumably if you have set it then you should know it already, if it is set dynamically you still have to have the name somewhere that you can access, then it is just a matter of storing it smartly, aka rather than using
my $x = \&foo;
Why not:
my %handler; $handler{'Bar::Foo'} = \&foo;
Now you can access the reference and have its name at the same time. The problem comes in with inheritance because maybe $y wants to call $x (foo) and thinks it is in Bar but it really is in Baz, does this mean it shouldn't be called, or just that it was inherited and it is ok to call?
Within foo, there is __PACKAGE__ which will give you Bar...
You may also want to have a look at Symbol Tables in perldoc perlmod....
Gurus, am I missing out on something?? ;-)
http://danconia.org
-- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
