请帮我明白了,为什么引用计数为每个级联的增长?我在过载子返回相同的对象,并期望,即引用计数将保持不变。但是,看来,Perl的克隆和存储对象的地方每次。为什么和我如何避免这种情况?
此外,我期待,该对象将在退出作用域之后摧毁,但由于非零引用计数销毁了只在全球毁灭的阶段。这看起来像一个内存泄漏。
#!/usr/bin/env perl
use strict;
use warnings;
use Devel::Refcount qw[refcount];
package AAA {
use Devel::Refcount qw[refcount];
use overload
'.' => sub {
print 'CONCAT, REFCOUNT: ', refcount( $_[0] ), "\n";
# return AAA->new;
return $_[0];
},
fallback => 0;
sub new { return bless {}, $_[0] }
sub DESTROY { print "DESTROY\n" }
}
print "--- start\n";
{
my $o = AAA->new;
my $s = '1' . ( '2' . ( '3' . ( '4' . ( '5' . $o ) ) ) );
print "--- exit scope\n";
print 'REFCOUNT: ', refcount($o), "\n";
}
print "--- end\n";
1;
下测试
产量
--- start
CONCAT, REFCOUNT: 1
CONCAT, REFCOUNT: 3
CONCAT, REFCOUNT: 5
CONCAT, REFCOUNT: 7
CONCAT, REFCOUNT: 9
--- exit scope
REFCOUNT: 6
--- end
DESTROY
就像延迟DESTROY
信息,添加弱裁判的对象指示的泄漏。泄漏似乎已经在Perl 5.28中引入。
use strict;
use warnings;
use Scalar::Util qw( weaken );
package AAA {
use overload
'.' => sub { $_[0] },
fallback => 0;
sub new { return bless {}, $_[0] }
sub DESTROY { print "DESTROY\n" }
}
my $w;
{
my $o = AAA->new;
weaken($w = $o);
my $s = '1' . ( '2' . ( '3' . ( '4' . ( '5' . $o ) ) ) );
print "Exiting scope...\n";
}
print "leak!\n" if defined($w);
#use Devel::Peek;
#Dump($w);
print "Global destruction...\n";
输出(前5.28):
Exiting scope...
DESTROY
Global destruction...
输出(5.28.0和5.28.1):
Exiting scope...
leak!
Global destruction...
DESTROY
请报告使用perlbug
命令行实用程序。
错误报告可以发现here。
它已被固定在5.30。这可能是固定在5.28.2。