short

You only deleted a stash entry pointing to the glob, not the glob which continues to exist as long as it is referenced in code.

long

A typeglob is a kind of hash with 6 fixed slots holding different data types² !

The scalar $n is the value referenced in the scalar slot of of the glob *n ¹

Think of a stash as a HoH  $pck{globname}{SCALAR} = \ 'value'

And like all Perl data *n is a C structure somewhere in memory and can be retrieved by its reference GLOBREF = \*n.

A symbol table is a hash holding names and references of such typeglobs.

Eg in your case

 %Foo:: = ( ..., "n"=> GLOBREF, ... )

The code print $n internally compiles to something like  print GLOBREF->{SCALAR} where GLOBREF was looked up in the stash at compile time and hardcoded into the OP-codes.

simplified analogy

globs and stashes are confusing, I tried to translate what is happening to HoHs.

Maybe that makes it clearer

BEGIN {
    # at compile time
    %stash =();                # 'package stash'
    $stash{n}{SCALAR} =undef;  # $n first use ( i.e. our-declaration )
    $n_glob = $stash{n};       # exists only hardcoded
}

# at runtime: execute compiled op-codes
$n_glob->{SCALAR}=123;       
delete $stash{n};
print $n_glob->{SCALAR};       # > 123

[download]

So deleting the Stash entry didn't delete the underlying glob (it's still referenced)

Update: You only sabotaged introspection at run time or further compilations with eval('print $n') .

Cheers Rolf

PS: Je suis Charlie!

update

¹) 

  DB<104> $n=123
 => 123

  DB<105> *n{SCALAR}
 => \123

[download]

²) i.e. refs to SCALAR, ARRAY, HASH, CODE, IO, GLOB, FORMAT see perlref

In reply to Re: What delete from symbol table really means? (Deleting typeglob of a specified package) by LanX
in thread What delete from symbol table really means? (Deleting typeglob of a specified package) by Discipulus

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