$A::{bbb} is not what you think it is:
$A::{bbb} = 3;
print $A::bbb;
__END__
Nothing. And
$A::{bbb} = 3;
$A::bbb = 4;
__END__
Modification of a read-only value attempted at - line 2.
Huh?
$A::bbb = 3;
$A::{bbb} = 4;
print "$_ => $A::{$_}\n" for keys %A::;
print $A::bbb,"\n";
__END__
bbb => *main::4
What's happening here? The %A:: has typeglobs as values, so assigning $A::{bbb} = 4 makes the literal 4 into a typeglob of the symbol table %main::. There's the glob, the typeglob entry in the symbol table, and the value in the glob's slot:
$A::bbb = 3;
print $A::{bbb},"\n";
print *A::bbb{SCALAR},"\n";
print ${*A::bbb{SCALAR}},"\n";
__END__
*A::bbb
SCALAR(0xa011750)
3
The special hash %A:: holds references to the typeglobs , these entries aren't the typeglobs themselves. but by deleting those entries you don't delete the glob. You mangle the lookup table only. their reference count must not go to zero. Since - as per your example - there's $A::bbb still present in compiled code, its reference count doesn't go to zero. So it is still accessible.
update: corrected above paragraph as per ysth's comment below. Thanks, ysth!
--shmem
_($_=" "x(1<<5)."?\n".q·/)Oo. G°\ /
/\_¯/(q /
---------------------------- \__(m.====·.(_("always off the crowd"))."·
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}
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