ForgotPasswordAgain has asked for the wisdom of the Perl Monks concerning the following question:

%SIG says "When a __DIE__ hook routine returns, the exception processing continues as it would have in the absence of the hook, unless the hook routine itself exits via a goto &sub, a loop exit, or a die()". Presumably "continues as it would have in the absence of the hook" means it dies, but what does "the hook routine itself exits via ... a loop exit" mean? Attempts: 
$ perl -Mstrict -wE'$SIG{__DIE__}=sub{last}; die "hi"'
Exiting subroutine via last at -e line 1.
Can't "last" outside a loop block at -e line 1.

# no strict
$ perl -wE'$SIG{__DIE__}=sub{last}; die "hi"'
Exiting subroutine via last at -e line 1.
Can't "last" outside a loop block at -e line 1.

# no warnings
$ perl -E'$SIG{__DIE__}=sub{last}; die "hi"'
Can't "last" outside a loop block at -e line 1.

# -e instead of -E
$ perl -e'$SIG{__DIE__}=sub{last}; die "hi"'
Can't "last" outside a loop block at -e line 1.

# put last in a while loop, dies
$ perl -e'$SIG{__DIE__}=sub{while(1){last}}; die "hi"'
hi at -e line 1.

# put last in a non-loop block, dies
$ perl -e'$SIG{__DIE__}=sub{{last}}; die "hi"'
hi at -e line 1.

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Edit: 
$ perl --version

This is perl 5, version 38, subversion 2 (v5.38.2) built for x86_64-ms
+ys-thread-multi

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Replies are listed 'Best First'.
Re: exit from $SIG{__DIE__} via a loop exit?
by choroba (Cardinal) on Mar 07, 2025 at 10:15 UTC
    The documentation was fixed 3 weeks ago.

    map{substr$_->[0],$_->[1]||0,1}[\*||{},3],[[]],[ref qr-1,-,-1],[{}],[sub{}^*ARGV,3]
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[d/l]
      Thanks, awesome coincidence, heh
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Re: exit from $SIG{__DIE__} via a loop exit?
by Corion (Patriarch) on Mar 07, 2025 at 06:48 UTC

    I thought you need a loop surrounding the die, but even that one complains instead of leaving the surrounding loop:

    #!perl
use 5.020;

$SIG{__DIE__}= sub { next };
for (1..2) {
    say $_;
    die;
}
say "done";

__END__
Can't "next" outside a loop block at tmp.pl line 3.

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    I remember that goto/next/last were mildly revamped to eliminate jumping into deeper blocks, and maybe this part of the documentation was not updated there...

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[d/l]
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Re: exit from $SIG{__DIE__} via a loop exit?
by ikegami (Patriarch) on Mar 07, 2025 at 15:48 UTC

    This is a loop exit: 

    sub f { last; }

say "before loop";

{
   say "start of loop";
   f();
   say "end of loop";
}

say "after loop";

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    ({ ... } is technically a loop. You can also use while/until loop, a foreach loop, or a C-style for loop.)

    This outputs 

    before loop
start of loop
Exiting subroutine via last at -e line 1.
after loop

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    So the documentation you quoted refers to programs such as the following: 
    $SIG{ __DIE__ } = sub { last; };

say "before loop";

{
   say "start of loop";
   die "!";
   say "end of loop";
}

say "after loop";

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    However, the code doesn't do what the documentation claims it does. 

    before loop
start of loop
Can't "last" outside a loop block at -e line 1.

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    choroba pointed out that the documentation was corrected a short while ago.

    However, the documentation is wrong, and it was corrected a short while ago. (Thanks to choroba for identifying this.)

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[d/l]
[select]
      Thanks, so at least I did understand what "exits via a loop exit" meant, but was led astray by the doc mismatch
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