__PACKAGE__ variable cannot use as replace of package name

chinaxing has asked for the wisdom of the Perl Monks concerning the following question:

I know the __PACKAGE__ will be compiled as package name at compile time. so why i cann't use it this way:

__PACKAGE__::bar()
[download]

in the Foo package's namespace?

this will cause the error :

Undefined subroutine &__PACAKGE__::bar called very thank you !

Comment on __PACKAGE__ variable cannot use as replace of package name Select or Download Code

Replies are listed 'Best First'.
Re: __PACKAGE__ variable cannot use as replace of package name by moritz (Cardinal) on Apr 23, 2013 at 07:24 UTC
`Undefined subroutine &__PACAKGE__::bar called` First of all you misspelled PACKAGE, which is not a good thing to do. But the real reason is that `__PACKAGE__` is only handled specially when it's a term on it own, not when it's part of a name. And finally `__PACKAGE__` as a separate term becomes a string, and you can't write `"MyPackage"::bar()` either. Perl 6 - the future is here, just unevenly distributed	[reply] [d/l] [select]
Re: __PACKAGE__ variable cannot use as replace of package name by kcott (Archbishop) on Apr 23, 2013 at 08:16 UTC
G'day chinaxing, I see ++moritz has described why you can't do this. Here's a way you might emulate the behaviour you're after. `$ perl -Mstrict -Mwarnings -E ' package Testonly; our $VERSION = "1.000"; { no strict "refs"; say "NO REFS: ", ${__PACKAGE__ . "::VERSION"}; } eval { say ${__PACKAGE__ . "::VERSION"}; }; warn "REFS: $@" if $@; ' NO REFS: 1.000 REFS: Can't use string ("Testonly::VERSION") as a SCALAR ref while "st +rict refs" in use at -e line 9.` [download] Note how I've used an anonymous namespace to isolate `${__PACKAGE__ . "::VERSION"}` in a scope that doesn't check for "strict refs". The eval that follows, demonstrates that the "`no strict "refs";`" is no longer in effect outside the anonymous namespace. -- Ken	[reply] [d/l] [select]
Re: __PACKAGE__ variable cannot use as replace of package name by choroba (Cardinal) on Apr 23, 2013 at 07:45 UTC
Crossposted at StackOverflow. It is considered polite to inform about crossposting so users not attending both sites do not waste their efforts on a problem already solved at the other corner of the Internet. لսႽ† ᥲᥒ⚪⟊Ⴙᘓᖇ Ꮅᘓᖇ⎱ Ⴙᥲ𝇋ƙᘓᖇ	[reply]