perl - 为什么这种魔术 Perl 内部变量的解决方法有效？

Question

t2.pl

1;
2;
3;

t.pl

package DB;

sub DB {
    print "HERE";

    our(undef, $f, undef) =  caller;

    # This does not work
    # my $ref =  \%{ "main::_<$f" };
    # $ref->{ 3 } =  {};

    # This does not work
    # *x =  $main::{ "_<$f" };
    # $x{ 3 } =  {};

    *dbline =  $main::{ "_<$f" };
    $dbline{ 3 } =  {};

    $DB::single =  0;
}

1;

PERL5DB="BEGIN{ require 't.pl' }" perl -d t2.pl

当您运行此代码时，您会得到 HEREHERE ，但是当您不使用 *dbline 为其别名时，神奇的 Perl 变量不起作用。 .

因此，当您将前两个示例更改为如下所示时:

*dbline =  $main::{ "_<$f" }; # <<<<<<<<< With this it works!!!!
*x =  $main::{ "_<$f" };
$x{ 3 } =  { };

断点开始工作。 (这适用于 Perl 5.14.4)

为什么它以这种方式工作？

DOC

Answer 1

这是reported前:

perldebguts says:

• Each hash %{"_<$filename"} contains breakpoints and actions keyed by line number. Individual entries (as opposed to the whole hash) are settable.

This implies that breakpoints set on %{"_<..."} apply to the named file. That is not actually true, as every %{"_<..."} hash sets breakpoints on lines in @DB::dbline, regardless of which file it refers to. The assumption is that debuggers will alias *DB::dbline to *{"_<..."} before setting any breakpoints.

Hence, all %{"_<..."} hashes are the same.

Is this a case where the documentation should be expanded to match the implementation? Or should we change the implementation to make each %{"_<..."} hash work on its corresponding @{"_<..."} array? The latter seems more useful to me.

在 5.20.0 中，行为是 changed这样你就不必别名 @DB::dbline :

$ PERL5DB='
  sub DB::DB {
      ($p,$f,$l) = caller;
      print "$f:$l\n";
      ${"::_<$f"}{3} = 1; # no need for alias
      $DB::single = 0
  }
' perl -d foo
foo:1
foo:3