Perl 中的引用是如何工作的 [英] How do references in Perl work

问题描述

谁能向我解释以下 Perl 代码中的 push 语句?我知道 perl 中的 push 是如何工作的，但我无法理解以下 push 命令中的第一个参数代表什么.我正在尝试解释某人的脚本.我试图 print "@a\n"; 但它只打印 ARRAY(0x9aa370) 这让我觉得推送没有做任何事情.任何帮助表示赞赏.谢谢！

我的@a = ();我的 $b = 10;我的 $c = 'a';推(@{$a[$b]}, $c);

解决方案

让我们分解一下.

@{...} 理解为 perlref 中的使用引用"

<块引用>

在您将标识符(或标识符链)作为变量或子例程名称的一部分放置的任何地方，您都可以用返回正确类型引用的 BLOCK 替换标识符.

那么 { ... } 块中的内容最好用于数组引用.您在那里有 $a[$b]，在索引 $b 处有一个 @a 元素，因此该元素必须是一个数组引用.

然后 @{...} 取消引用它并 push es 一个新元素 $c 给它.总之，$c 被复制到一个匿名数组的(唯一)元素中，该元素的引用位于数组 @a 的索引 $b.

还有一个关键部分:因为实际上没有 arrayref ^†，autovivification 启动并创建.由于 $b 之前的索引处没有元素，它们也被创建，值为 undef.

现在请完成

• 教程 perlreftut 和

• 数据结构手册perldsc

同时使用 perlref 链接在开头以获得完整参考.

<小时>

对于复杂的数据结构，能够看到它们很有用，并且有工具可以做到这一点.最常用的是核心Data::Dumper，这里有一个例子使用 Data::Dump

perl -MData::Dump=dd -wE'@ary = (1);push @{$ary[3]}, "啊";dd \@ary'

带输出

<前>[1, undef, undef, ["啊"]]

其中 [] 里面表示一个 arrayref，它的唯一元素是字符串 ah.

<小时>

^† 更准确地说，一个 undef 标量被取消引用，因为这发生在 lvalue 上下文中，自动激活就开始了.感谢 ikegami 的评论.例如，请参阅这篇文章及其链接.

Can anyone explain the push statement in the following Perl code to me? I know how push in perl works but I can't understand what the first argument in following push command represents. I am trying to interpret someone's script. I tried to print "@a\n"; but it only printed ARRAY(0x9aa370) which makes me think that the push is not doing anything. Any help is appreciated. Thanks!

my @a = (); 
my $b = 10;
my $c = 'a';
push(@{$a[$b]}, $c);

解决方案

Let's break it down.

The @{...} is understood from "Using References" in perlref

Anywhere you'd put an identifier (or chain of identifiers) as part of a variable or subroutine name, you can replace the identifier with a BLOCK returning a reference of the correct type.

So what is inside { ... } block had better work out to an array reference. You have $a[$b] there, an element of @a at index $b, so that element must be an arrayref.

Then @{...} dereferences it and pushes a new element $c to it. Altogether, $c is copied into a (sole) element of an anonymous array whose reference is at index $b of the array @a.

And a crucial part: as there is in fact no arrayref ^† there, the autovivification kicks in and it is created. Since there are no elements at indices preceding $b they are created as well, with value undef.

Now please work through

• tutorial perlreftut and

• data-structures cookbook perldsc

while using perlref linked in the beginning for a full reference.

---

With complex data structures it is useful to be able to see them, and there are tools for that. A most often used one is the core Data::Dumper, and here is an example with Data::Dump

perl -MData::Dump=dd -wE'@ary = (1); push @{$ary[3]}, "ah"; dd \@ary'

with output

[1, undef, undef, ["ah"]]

where [] inside indicate an arrayref, with its sole element being the string ah.

---

^† More precisely, an undef scalar is dereferenced and since this happens in an lvalue context the autovivification goes. Thanks to ikegami for a comment. See for instance this post with its links.

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