Re: my \&foo = sub { }?

[email protected] (Aristotle Pagaltzis via perl5-porters) Sat, 18 Apr 2026 04:53:37 +0200
Newsgroups perl.perl5.porters
Message-ID <[email protected]>
* Eric Brine <[email protected]> [2026-04-08 01:41]:
> When using the lexical_subs, refaliasing and declared_refs features,
>
> We have C<< my sub foo; \&foo = REF; >>
> We have C<< my \$foo = REF; >>
> Why don't we have C<< my \&foo = REF; >>

Because we don’t have C<< my &foo >>.