$SIG{__DIE__} handler doesn't work when called explicitly

[mauke]

Description

If you register a subroutine as a $SIG{__DIE__} handler, then call it normally, and that sub then (directly or indirectly) throws an exception, the $SIG{__DIE__} behavior doesn't kick in; i.e. the sub is not re-invoked to handle the exception.

Steps to Reproduce

$ perl -we 'my $x = 0; sub foo { print "foo()\n"; die $x++ } $SIG{__DIE__} = \&foo; foo'
foo()
0 at -e line 1.

(Note that $SIG{__DIE__} has no effect here.)

Expected behavior

$ perl -we 'my $x = 0; sub foo { print "foo()\n"; die $x++ } $SIG{__DIE__} = \&foo; foo'
foo()
foo()
1 at -e line 1.

There should be exactly two calls to foo: One from foo in the main code, the other via $SIG{__DIE__}. It should not recurse further because as perldoc -v %SIG explains:

The __DIE__ handler is explicitly disabled during the call, so that you can die from a __DIE__ handler.

[demerphq]

There should be exactly two calls to foo

Agreed, with the minor quibble that if you changed the code to

perl -we 'my $x = 0; sub foo { print "foo()\n"; die $x++ } $SIG{__DIE__} = \&foo; $SIG{__DIE__}->()'

I think I would expect that foo is called only once. Although I could be convinced it should be called twice. Regardless we should document this explicitly.