Arrays 引用数组的大小已下降到一个元素

亲爱的perl程序员同事：

我想访问此阵列

my @vsrvAttribs = qw(
  Code
  Description
  vsrv_id
  vsrv_name
  vsrv_vcpu_no
  vsrv_vmem_size
  vsrv_vdspace_alloc
  vsrv_mgmt_ip
  vsrv_os
  vsrv_virt_platf
  vsrv_owner
  vsrv_contact
  vsrv_state
);

通过一个由变量和字符串后缀组成的变量，这当然会导致如下错误消息

Can't use string ("@vsrvAttribs") as an ARRAY ref while "strict refs" in use at cmdbuild.pl line 262.`

因此，我决定通过散列获取对数组的引用

my %attribs = ( vsrv => @vsrvAttribs );

这是我需要获取上述数组内容的代码

foreach my $classTypeKey (keys %classTypes) {
  my @attribs = $attribs{$classTypeKey};
  print Dumper(\@attribs);
}

看起来我可以得到对数组@vsrvAttribs的引用，但是当我用Dumper检查数组的内容时，数组只有一个元素

$VAR1 = [
          'Code'
        ];

您知道问题出在哪里吗？

如何将数组存储在散列中并在以后访问它？ 您需要通过引用存储数组，如下所示：

my %attribs = ( vsrv => \@vsrvAttribs );

foreach my $classTypeKey (keys %classTypes) {
  # make a copy of the array by dereferencing
  my @attribs = @{ $attribs{$classTypeKey} };

  # OR just use the array reference if profiling shows performance issues:
  my $attribs = $attribs{$classTypeKey}

  # these will show the same thing if you haven't done anything to @attribs
  # in the interim
  print Dumper(\@attribs);
  print Dumper($attribs);
}

$VAR1 = {
          'vsrv_contact' => 'vsrv_state',
          'vsrv_virt_platf' => 'vsrv_owner',
          'vsrv' => 'Code',
          'vsrv_name' => 'vsrv_vcpu_no',
          'vsrv_mgmt_ip' => 'vsrv_os',
          'Description' => 'vsrv_id',
          'vsrv_vmem_size' => 'vsrv_vdspace_alloc'
        };

注意

@

符号前的反斜杠。这告诉perl您需要对数组的引用

然后，当访问存储在

$attribs{vsrv}

中的数组时，需要将其视为引用，而不是数组。您将执行以下操作：

my %attribs = ( vsrv => \@vsrvAttribs );

foreach my $classTypeKey (keys %classTypes) {
  # make a copy of the array by dereferencing
  my @attribs = @{ $attribs{$classTypeKey} };

  # OR just use the array reference if profiling shows performance issues:
  my $attribs = $attribs{$classTypeKey}

  # these will show the same thing if you haven't done anything to @attribs
  # in the interim
  print Dumper(\@attribs);
  print Dumper($attribs);
}

$VAR1 = {
          'vsrv_contact' => 'vsrv_state',
          'vsrv_virt_platf' => 'vsrv_owner',
          'vsrv' => 'Code',
          'vsrv_name' => 'vsrv_vcpu_no',
          'vsrv_mgmt_ip' => 'vsrv_os',
          'Description' => 'vsrv_id',
          'vsrv_vmem_size' => 'vsrv_vdspace_alloc'
        };

为什么只得到一个值，而数组的其余部分又到哪里去了？

@vsrvAttribs

中缺少的值并没有丢失，而是作为键和值分配给了

%attribs

本身。在完成作业后，尝试添加以下内容，您将亲自看到：

my %attribs = ( vsrv => @vsrvAttribs );
print Dumper(\%attribs);

您将看到如下输出：

my %attribs = ( vsrv => \@vsrvAttribs );

foreach my $classTypeKey (keys %classTypes) {
  # make a copy of the array by dereferencing
  my @attribs = @{ $attribs{$classTypeKey} };

  # OR just use the array reference if profiling shows performance issues:
  my $attribs = $attribs{$classTypeKey}

  # these will show the same thing if you haven't done anything to @attribs
  # in the interim
  print Dumper(\@attribs);
  print Dumper($attribs);
}

$VAR1 = {
          'vsrv_contact' => 'vsrv_state',
          'vsrv_virt_platf' => 'vsrv_owner',
          'vsrv' => 'Code',
          'vsrv_name' => 'vsrv_vcpu_no',
          'vsrv_mgmt_ip' => 'vsrv_os',
          'Description' => 'vsrv_id',
          'vsrv_vmem_size' => 'vsrv_vdspace_alloc'
        };

这是因为perl通过将内容

@vsrvAttribs

扩展为列表文本

（）

的多个参数来解释赋值：

这在perl中是合法的，您可能有理由这样做，但在您的情况下，这并不是您想要的

顺便说一句，如果数组中有偶数个元素，您可能会被警告perl正在做一些您可能不想做的事情。你的13个元素使散列键“vsrv”变为14，这是偶数。Perl将获取任何元素数为偶数的列表，并愉快地将其转换为散列。如果您的数组中有另一个包含15个元素和哈希键的元素，您将得到一个警告：

foo.pl第28行哈希赋值中的奇数元素。

有关详细信息，请参阅中的“生成引用”和“使用引用”。

如何将数组存储在哈希中，并在以后访问它？ 您需要通过引用存储数组，如下所示：

my %attribs = ( vsrv => \@vsrvAttribs );

foreach my $classTypeKey (keys %classTypes) {
  # make a copy of the array by dereferencing
  my @attribs = @{ $attribs{$classTypeKey} };

  # OR just use the array reference if profiling shows performance issues:
  my $attribs = $attribs{$classTypeKey}

  # these will show the same thing if you haven't done anything to @attribs
  # in the interim
  print Dumper(\@attribs);
  print Dumper($attribs);
}

$VAR1 = {
          'vsrv_contact' => 'vsrv_state',
          'vsrv_virt_platf' => 'vsrv_owner',
          'vsrv' => 'Code',
          'vsrv_name' => 'vsrv_vcpu_no',
          'vsrv_mgmt_ip' => 'vsrv_os',
          'Description' => 'vsrv_id',
          'vsrv_vmem_size' => 'vsrv_vdspace_alloc'
        };

注意

@

符号前的反斜杠。这告诉perl您需要对数组的引用

然后，当访问存储在

$attribs{vsrv}

中的数组时，需要将其视为引用，而不是数组。您将执行以下操作：

my %attribs = ( vsrv => \@vsrvAttribs );

foreach my $classTypeKey (keys %classTypes) {
  # make a copy of the array by dereferencing
  my @attribs = @{ $attribs{$classTypeKey} };

  # OR just use the array reference if profiling shows performance issues:
  my $attribs = $attribs{$classTypeKey}

  # these will show the same thing if you haven't done anything to @attribs
  # in the interim
  print Dumper(\@attribs);
  print Dumper($attribs);
}

$VAR1 = {
          'vsrv_contact' => 'vsrv_state',
          'vsrv_virt_platf' => 'vsrv_owner',
          'vsrv' => 'Code',
          'vsrv_name' => 'vsrv_vcpu_no',
          'vsrv_mgmt_ip' => 'vsrv_os',
          'Description' => 'vsrv_id',
          'vsrv_vmem_size' => 'vsrv_vdspace_alloc'
        };

为什么只得到一个值，而数组的其余部分又到哪里去了？

@vsrvAttribs

中缺少的值并没有丢失，而是作为键和值分配给了

%attribs

本身。在完成作业后，尝试添加以下内容，您将亲自看到：

my %attribs = ( vsrv => @vsrvAttribs );
print Dumper(\%attribs);

您将看到如下输出：

my %attribs = ( vsrv => \@vsrvAttribs );

foreach my $classTypeKey (keys %classTypes) {
  # make a copy of the array by dereferencing
  my @attribs = @{ $attribs{$classTypeKey} };

  # OR just use the array reference if profiling shows performance issues:
  my $attribs = $attribs{$classTypeKey}

  # these will show the same thing if you haven't done anything to @attribs
  # in the interim
  print Dumper(\@attribs);
  print Dumper($attribs);
}

$VAR1 = {
          'vsrv_contact' => 'vsrv_state',
          'vsrv_virt_platf' => 'vsrv_owner',
          'vsrv' => 'Code',
          'vsrv_name' => 'vsrv_vcpu_no',
          'vsrv_mgmt_ip' => 'vsrv_os',
          'Description' => 'vsrv_id',
          'vsrv_vmem_size' => 'vsrv_vdspace_alloc'
        };

这是因为perl通过将内容

@vsrvAttribs

扩展为列表文本

（）

的多个参数来解释赋值：

这在perl中是合法的，您可能有理由这样做，但在您的情况下，这并不是您想要的

顺便说一句，如果数组中有偶数个元素，您可能会被警告perl正在做一些您可能不想做的事情。你的13个元素使散列键“vsrv”变为14，这是偶数。Perl将获取任何元素数为偶数的列表，并愉快地将其转换为散列。如果您的数组中有另一个包含15个元素和哈希键的元素，您将得到一个警告：

foo.pl第28行哈希赋值中的奇数元素。

---

有关详细信息，请参阅中的“生成引用”和“使用引用”。

如果在哈希定义中使用空数组，如

my %attribs = ( vsrv => @vsrv_attribs )

数组被扩展并用作键/值对，因此

我的%attribs=( vsrv=>“代码”， Description=>'vsrv_id'， vsrv_名称=>“vsrv_vcpu_编号”， vsrv_vmem_size=>“vsrv_VDU空间_alloc”， ... )

Perl散列元素的值只能是标量值，因此，如果需要一个值数组，则必须获取一个引用，如下所示

在Perl标识符中使用大写字母表示除全局字符以外的任何内容（如包名）也是一个坏主意。本地名称是传统的小写字母数字加下划线，因此是

$class\u type\u key

而不是

$classTypeKey

use strict;
use warnings;

use Data::Dumper;

my @vsrv_attribs = qw(
  Code
  Description
  vsrv_id
  vsrv_name
  vsrv_vcpu_no
  vsrv_vmem_size
  vsrv_vdspace_alloc
  vsrv_mgmt_ip
  vsrv_os
  vsrv_virt_platf
  vsrv_owner
  vsrv_contact
  vsrv_state
);

my %attribs = (
  vsrc => \@vsrv_attribs,
);


for my $class_type_key (keys %attribs) {
  my $attribs = $attribs{$class_type_key};
  print Dumper $attribs;
}

输出

$VAR1 = [
          'Code',
          'Description',
          'vsrv_id',
          'vsrv_name',
          'vsrv_vcpu_no',
          'vsrv_vmem_size',
          'vsrv_vdspace_alloc',
          'vsrv_mgmt_ip',
          'vsrv_os',
          'vsrv_virt_platf',
          'vsrv_owner',
          'vsrv_contact',
          'vsrv_state'
        ];

---

如果在散列定义中使用裸数组，如

my %attribs = ( vsrv => @vsrv_attribs )

数组被扩展并用作键/值对，因此

我的%attribs=( vsrv=>“代码”， Description=>'vsrv_id'， vsrv_名称=>“vsrv_vcpu_编号”， vsrv_vmem_size=>“vsrv_VDU空间_alloc”， ... )

Perl散列元素的值只能是标量值，因此，如果需要一个值数组，则必须获取一个引用，如下所示

在Perl标识符中使用大写字母表示除全局字符以外的任何内容（如包名）也是一个坏主意。本地名称是传统的小写字母数字加下划线，因此是

$class\u type\u key

而不是

$classTypeKey

use strict;
use warnings;

use Data::Dumper;

my @vsrv_attribs = qw(
  Code
  Description
  vsrv_id
  vsrv_name
  vsrv_vcpu_no
  vsrv_vmem_size
  vsrv_vdspace_alloc
  vsrv_mgmt_ip
  vsrv_os
  vsrv_virt_platf
  vsrv_owner
  vsrv_contact
  vsrv_state
);

my %attribs = (
  vsrc => \@vsrv_attribs,
);


for my $class_type_key (keys %attribs) {
  my $attribs = $attribs{$class_type_key};
  print Dumper $attribs;
}

输出

$VAR1 = [
          'Code',
          'Description',
          'vsrv_id',
          'vsrv_name',
          'vsrv_vcpu_no',
          'vsrv_vmem_size',
          'vsrv_vdspace_alloc',
          'vsrv_mgmt_ip',
          'vsrv_os',
          'vsrv_virt_platf',
          'vsrv_owner',
          'vsrv_contact',
          'vsrv_state'
        ];

---

你到底想解决什么问题？解释你想要实现什么，而不是如何实现。你想要解决的确切问题是什么？解释你想要实现什么，而不是如何实现。是的，你是对的。我已经弄明白了，但你的反应比我回答我的问题更快。无论如何，谢谢你。是的，你说得对。我已经弄明白了，但你的反应比我回答我的问题更快。无论如何，谢谢你。博罗丁，谢谢你宝贵的回答和非常好的解释。作为初学者，这对我很有帮助。博罗丁，谢谢你宝贵的回答和很好的解释。对我来说