揭秘 Perl glob (*)

本文介绍了揭秘 Perl glob (*)的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

在这个问题中，海报询问了如何在一行中执行以下操作:

sub my_sub {我的 $ref_array = shift;我的@array = @$ref_array;}

根据我对基本 Perl 魔法的了解，我可以通过简单地使用以下内容来避免:

sub my_sub {我的 $ref_array = shift;对于(@$ref_array){#在这里用 $_ 做点什么};#在此处使用 $ref_array->[$element]}

但是在 这个答案中，SO 的一位当地僧侣 tchrist 建议:

sub my_sub {本地 *array = shift();#在这里使用@array}

当我问

在尝试学习中级 Perl魔法，我能问一下，你是什么在这里设置什么?你是将@array 的引用设置为已经传入的arrayref?如何你知道你创建了@array 和不是 %array 或 $array?我可以在哪里了解有关此 * 运算符的更多信息(perlop?).谢谢！

有人建议我将其作为新帖子提问，尽管他确实提供了很好的参考.无论如何，来了?有人可以解释什么被分配给什么以及如何创建@array 而不是 %array 或 $array?谢谢.

解决方案

分配给 glob

*glob = VALUE

包含一些依赖于 VALUE 类型的魔法(即，返回值，比如 Scalar::Util::reftype(VALUE)).如果VALUE 是对标量、数组、散列或子例程的引用，则仅符号表中的该条目将被覆盖.

这个成语

local *array = shift();#在这里使用@array

当子例程的第一个参数是数组引用时，按文档中的说明工作.如果第一个参数是标量引用，那么只有 $array 而不是 @array 会受到赋值的影响.

一个小演示脚本，看看发生了什么:

无严格；子F{本地*数组=移位；打印 "@array = @array ";打印 "$array = $array ";打印 "\%array = ",%array," ";打印 "------------------- ";}$array = "原始标量";%array = ("原始" => "哈希");@array = ("原始","数组");$foo = "foo";@foo = ("foo","bar");%foo = ("FOO" => "foo");F ["新建","数组"];# 数组引用F "新标量";# 标量参考F {新" =>哈希"};# 哈希引用F *富;# typeglobF'富';# 不是引用，而是分配变量的名称F '别的东西';# 不是参考F ();# 未定义

输出:

@array = 新数组$array = 原始标量%array = 原始哈希------------------@array = 原始数组$array = 新标量%array = 原始哈希------------------@array = 原始数组$array = 原始标量%array = 新哈希------------------@array = foo 栏$数组 = foo%array = FOOfoo------------------@array = foo 栏$数组 = foo%array = FOOfoo------------------@数组 =$数组 =%数组 =------------------@array = 原始数组$array = 原始标量%array = 原始哈希------------------

附加文档位于 perlmod 和 perldata.早在引用成为 Perl 的一部分之前，这个习惯用法有助于将数组和散列传递到子例程中.

In this question the poster asked how to do the following in one line:

sub my_sub { my $ref_array = shift; my @array = @$ref_array; }

which with my knowledge of the basic Perl magic I would avoid by simply using something like:

sub my_sub { my $ref_array = shift; for (@$ref_array) { #do somthing with $_ here }; #use $ref_array->[$element] here }

However in this answer one of SO's local monks tchrist suggested:

sub my_sub { local *array = shift(); #use @array here }

When I asked

In trying to learn the mid-level Perl magic, can I ask, what is it that you are setting to what here? Are you setting a reference to @array to the arrayref that has been passed in? How do you know that you create @array and not %array or $array? Where can I learn more about this * operator (perlop?). Thanks!

I was suggested to ask it as a new post, though he did give nice references. Anyway, here goes? Can someone please explain what gets assigned to what and how come @array gets created rather than perhaps %array or $array? Thanks.

解决方案

Assignment to a glob

*glob = VALUE

contains some magic that depends on the type of VALUE (i.e., return value of, say, Scalar::Util::reftype(VALUE)). If VALUE is a reference to a scalar, array, hash, or subroutine, then only that entry in the symbol table will be overwritten.

This idiom

local *array = shift(); #use @array here

works as documented when the first argument to the subroutine is an array reference. If the first argument was instead, say, a scalar reference, then only $array and not @array would be affected by the assignment.

A little demo script to see what is going on:

no strict; sub F { local *array = shift; print "@array = @array "; print "$array = $array "; print "\%array = ",%array," "; print "------------------ "; } $array = "original scalar"; %array = ("original" => "hash"); @array = ("orignal","array"); $foo = "foo"; @foo = ("foo","bar"); %foo = ("FOO" => "foo"); F ["new","array"]; # array reference F "new scalar"; # scalar reference F {"new" => "hash"}; # hash reference F *foo; # typeglob F 'foo'; # not a reference, but name of assigned variable F 'something else'; # not a reference F (); # undef

Output:

@array = new array $array = original scalar %array = originalhash ------------------ @array = orignal array $array = new scalar %array = originalhash ------------------ @array = orignal array $array = original scalar %array = newhash ------------------ @array = foo bar $array = foo %array = FOOfoo ------------------ @array = foo bar $array = foo %array = FOOfoo ------------------ @array = $array = %array = ------------------ @array = orignal array $array = original scalar %array = originalhash ------------------

Additional doc at perlmod and perldata. Back in the days before references were a part of Perl, this idiom was helpful for passing arrays and hashes into subroutines.