Hi D', I agree whatever you are saying. But i need to support somebody's code without annoying users by breaking their old code which depends on this interface.
Yes, i can expand the scope of the routine now and let users pass in the additional information. But again if i can guess it somehow, why to make it mandatory in the interface. -Sharad > -----Original Message----- > From: Wiggins d'Anconia [mailto:[EMAIL PROTECTED] > Sent: Wednesday, October 22, 2003 8:53 PM > To: Gupta, Sharad > Cc: [EMAIL PROTECTED] > Subject: Re: FQN from references > > > Gupta, Sharad wrote: > > 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]
