Assuming you're using Exporter or something like that, then the fact is that foo is in the current namespace, the one it was imported into. If you want to know where it came from, you'd probably have to dink with the import function to record the import action somehow.

Another possibility (not necessarily recommended) is to add a "secret" protocol to the foo function whereby it returns its native package e.g. when passed a certain argument:

sub foo {
    $_[-1] eq '__WHENCE__' and return __PACKAGE__;
    . . . .
}
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caller knows the package name (show below) at run-time. Is there a way (B:: modules?) to extract it without calling the function?

<Reveal this spoiler or all in this thread>

Maybe Sub::Identify can help too.

use Sub::Identify 'stash_name';
my $package = stash_name( $some_coderef );
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Frankly, I get tired of always trying to figure out stuff about subroutines (what's their name, if anonymous? What's their code? What package are they from? etc.). So I bundled all of that up in Sub::Information:

use Sub::Information;
print inspect(\&coderef)->package;
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if (\&foo == \&sourcea::foo) {
  ...
} elsif (\&foo == \&sourceb::foo) {
  ...
} else {
  die "What? The foo() comes from some other package!"
}
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