Perl：哈希引用访问键数组 [英] Perl: Hash ref accessing array of keys

问题描述

我有一个似乎很基本的问题，但我无法弄清楚。假设我在Perl中有一个哈希引用。我想通过一组键来获取一组值。

I have a question that seems basic, but I can't figure out. Say that I have a hash reference in Perl. I want to get an array of values through an array of keys.

以下是它如何使用散列，而不是散列引用：

Here's how it'd work with a hash, not a hash reference:

my %testHash = ( "a" => 1, "b" => 2, "c" => 3 );
my @testKeys = ("a", "b", "c");

my @testValues = @testHash{@testKeys};

现在假设我有一个散列引用，

Now suppose I have a hash reference,

my $hashRef = {"a" => 1, "b" => 2, "c" => 3};

我尝试了以下两种方式：

I've tried the following two ways:

my @values = @{$hashRef->{@testKeys}};
my @values = $hashRef->{@testKeys};

但这两个都不正确。有没有一个正确的方法，或者我只需要在每次我想要这样做时都取消引用散列ref？

But neither is correct. Is there a correct way, or do I just have to dereference the hash ref every time I want to do this?

推荐答案

您'关闭：

my @values = @$hashref{@testKeys};     ## (1)

或

my @values = @{$hashref}{@testKeys};   ## (2)

更多示例请参见切片perllol文档。

在Perlref文档中使用引用给出了一般规则。

"Using References" in the perlref documentation gives the general rules.

1。无论你在哪里放置一个标识符（或者标识符链）作为变量或子程序名称的一部分，都可以用包含正确类型引用的简单标量变量替换标识符。

1. 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 simple scalar variable containing a reference of the correct type.

这解释了为什么（1）有效：用简单的标量 $替换了标识符 testHash hashRef 。

This explains why (1) works: you replaced the identifier testHash with the simple scalar $hashRef.

2。无论您将标识符（或标识符链）作为变量或子例程名称的一部分放置在任何位置，都可以使用BLOCK来替换标识符，以返回正确类型的引用。

2. 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.

上面的代码片段（2）几乎相同，但语法较为庞大。代替标识符 testHash ，您可以编写一个返回散列引用的块， ie ， {$ hashRef} 。

Snippet (2) above is nearly the same but has a little bulkier syntax. In place of the identifier testHash, you write a block returning a reference to a hash, i.e., {$hashRef}.

这里的大括号包含一个善意的块，所以您可以计算并返回一个引用，如

The braces here enclose a bona fide block, so you can compute and return a reference, as in

push @{ $cond ? \@a1 : \@a2 }, "some value";

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