perl – 嵌套闭包和捕获的变量

概述我有这个带有嵌套闭包的例子,它演示了内存泄漏 use v5.10;use strict;package Awesome;sub new { bless {steps => [], surprise => undef}, shift;}sub say { print "awesome: ", $_[1], "\n";}sub prepare { my ( 我有这个带有嵌套闭包的例子,它演示了内存泄漏

use v5.10;use strict;package Awesome;sub new {    bless {steps => [],surprise => undef},shift;}sub say {    print "awesome: ",$_[1],"\n";}sub prepare {    my ($self,@steps) = @_;    for my $s (@steps) {        push @{$self->{steps}},sub {            $self->say($s);            if ($s eq 'pony') {                $self->{surprise} = sub {                    $s;                }            }        };    }}sub make {    my $self = shift;    while (my $step = shift @{$self->{steps}}) {        $step->();    }    if ($self->{surprise}) {        printf("And you have surprise: %s\n",$self->{surprise}->());    }}sub DESTROY {    warn "destroying";}package main;my $a = Awesome->new;$a->prepare('barbIE','pony','flash');$a->make();

我的perl输出是

awesome: barbIEawesome: ponyawesome: flashAnd you have surprise: ponydestroying at /tmp/t.pl line 43 during global destruction.

而这个“在全局破坏期间”意味着物体不能以正常方式被破坏,因为它有一些循环引用.

但是,唯一的循环引用是由

push @{$self->{steps}},sub {            $self->say($s);

我们在第一次关闭时使用$self.然后在make()里面我们将删除这些步骤和循环引用.但看起来这个嵌套的封闭与“惊喜”会产生问题.例如,如果我们不将“pony”传递给prepare(),那么输出将如预期的那样好：

awesome: barbIEawesome: flashdestroying at /tmp/t.pl line 43.

那么,perl中的嵌套闭包是否捕获了与已经捕获的上层闭包相同的变量,即使我们没有使用它们？

解决方法 Perl过去常常在嵌套的闭包中过度捕获,但自5.18以来它不会这样做.

$tail -n 9 a.pl   # ModifIEd to make clearer when the object is destroyed.package main;{   my $a = Awesome->new;   $a->prepare('barbIE','flash');   $a->make();}print "done.\n";

 

.16.3t/bin/perl a.plawesome: barbIEawesome: ponyawesome: flashAnd you have surprise: ponydone.destroying at a.pl line 43 during global destruction.

 

.18.2t/bin/perl a.plawesome: barbIEawesome: ponyawesome: flashAnd you have surprise: ponydestroying at a.pl line 43.done.

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