perl:打印对象属性 [英] perl: printing object properties
问题描述
我正在使用 Net::Amazon::EC2 库，但找不到打印对象属性的简单方法:
I'm playing a bit with the Net::Amazon::EC2 libraries, and can't find out a simple way to print object properties:
这有效:
my $snaps = $ec2->describe_snapshots();
foreach my $snap ( @$snaps ) {
print $snap->snapshot_id . " " . $snap->volume_id . "\n";
}
但如果我尝试:
print "$snap->snapshot_id $snap->volume_id \n";
我明白了
Net::Amazon::EC2::Snapshot=HASH(0x4c1be90)->snapshot_id
是否有一种简单的方法可以在打印件中打印属性的值?
Is there a simple way to print the value of the property inside a print?
推荐答案
不是你想要的方式.事实上，您对 $snap->snapshot_id 所做的就是调用一个方法(如在 sub 中那样).Perl 不能在双引号字符串中做到这一点.它将插入您的变量 $snap.这变成了类似于 HASH(0x1234567) 的东西，因为它是这样的:散列的 bless 引用.
Not in the way you want to do it. In fact, what you're doing with $snap->snapshot_id is calling a method (as in sub). Perl cannot do that inside a double-quoted string. It will interpolate your variable $snap. That becomes something like HASH(0x1234567) because that is what it is: a blessed reference of a hash.
插值仅适用于标量(和数组，但我会忽略).你可以去:
The interpolation only works with scalars (and arrays, but I'll omit that). You can go:
print "$foo $bar"; # scalar
print "$hash->{key}"; # scalar inside a hashref
print "$hash->{key}->{moreKeys}->[0]"; # scalar in an array ref in a hashref...
不过，有一种方法可以做到:您可以在带引号的字符串中引用和取消引用它，就像我在这里所做的那样:
There is one way to do it, though: You can reference and dereference it inside the quoted string, like I do here:
use DateTime;
my $dt = DateTime->now();
print "${\$dt->epoch }"; # both these
print "@{[$dt->epoch]}"; # examples work
但这看起来很丑，所以我不推荐.请改用第一种方法！
But that looks rather ugly, so I would not recommend it. Use your first approach instead!
如果您仍然对其工作原理感兴趣，您可能还想查看这些 Perl 常见问题解答:
If you're still interested in how it works, you might also want to look at these Perl FAQs:
来自 perlref:
这里有一个将子程序调用插入到字符串中的技巧:
Here's a trick for interpolating a subroutine call into a string:
print "My sub returned @{[mysub(1,2,3)]} that time.\n";
它的工作方式是当@{...} 出现在双引号中时字符串，它被评估为一个块.该块创建对一个的引用包含调用 mysub(1,2,3) 的结果的匿名数组.所以整个块返回一个对数组的引用，然后被@{...} 取消引用并卡在双引号字符串中.这Chicanery 对任意表达式也很有用:
The way it works is that when the @{...} is seen in the double-quoted string, it's evaluated as a block. The block creates a reference to an anonymous array containing the results of the call to mysub(1,2,3) . So the whole block returns a reference to an array, which is then dereferenced by @{...} and stuck into the double-quoted string. This chicanery is also useful for arbitrary expressions:
print "That yields @{[$n + 5]} widgets\n";
同样，返回标量引用的表达式可以是通过 ${...} 取消引用.因此，上面的表达式可以写成如:
Similarly, an expression that returns a reference to a scalar can be dereferenced via ${...} . Thus, the above expression may be written as:
print "That yields ${\($n + 5)} widgets\n";
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