Question

我有一个Perl脚本试图将一些已配置的DateTime和DateTime::Duration个实例设置为Readonly个常量。但是，如果它们是Readonly，我会在尝试对这些对象进行数学运算时看到奇怪的行为。这是一个最小的例子：

#!/usr/bin/perl -w

use strict;
use warnings;

use DateTime;
use Readonly;

Readonly my $X => DateTime->now;
my $x = DateTime->now;

Readonly my $Y => DateTime::Duration->new( days => 3 );
my $y = DateTime::Duration->new( days => 3 );

my $a = $X - $Y;
my $b = $x - $y;

print "$a\n";
print "$b\n";

在我的系统上（OSX上的Perl 5.10.0）显示：

$ ./datetime_test.pl 
Argument "2011-07-12T20:36:08" isn't numeric in subtraction (-) at ./datetime_test.pl line 15.
-4305941629
2011-07-09T20:36:08

所以看起来好像使DateTime和DateTime::Duration Readonly导致它们无法正常运行。这是一个错误吗？或者我使用Readonly错了？我也尝试了Readonly::Scalar和Readonly::Scalar1，两者的行为方式相同。

Answer 1

问题在于它们是对象（引用），而不是普通的标量。您需要Readonly引用中包含的值，而不是引用本身;但事实证明这很棘手。这样的事情似乎有效：

use Readonly;
use DateTime;

# you can't just say "Readonly %$dt"; here at least, it dies on blessed refs
sub makeRO {
  my $dt = shift;
  while (my ($k, $v) = each %$dt) {
    Readonly $dt->{$k} => $v;
  }
}

my $x = DateTime::Duration->new(days => 3);
makeRO($x);
my $y = DateTime::Duration->new(days => 3);

my $a = $x - $y;
# print "$a\n"; # this isn't overloaded; you'll get "DateTime::Duration=HASH(...)"
print $a->days, "\n";

如何在Perl中创建Readonly DateTime和DateTime :: Duration

1 个答案: