检查ruby中实例的类型

Question

我正在调试一个应用程序，在这个应用程序中，我需要检查DNS查询的资源类是否为IN类型。以下是我的发现：

[6] pry(#<DNS::Server>)> resource_class
=> Resolv::DNS::Resource::IN::A
[7] pry(#<DNS::Server>)> resource_class.inspect
=> "Resolv::DNS::Resource::IN::A"
[8] pry(#<DNS::Server>)> resource_class.class.name
=> "Class"
[9] pry(#<DNS::Server>)> resource_class.kind_of? Class
=> true
[10] pry(#<DNS::Server>)> resource_class.kind_of? Resolv::DNS::Resource::IN::A
=> false
[11] pry(#<DNS::Server>)> resource_class.instance_of? Class
=> true
[12] pry(#<DNS::Server>)> resource_class.instance_of? Resolv::DNS::Resource::IN::A
=> false
[13] pry(#<DNS::Server>)> resource_class.is_a? Class
=> true
[14] pry(#<DNS::Server>)> resource_class.is_a? Resolv::DNS::Resource::IN::A
=> false

我试图使用resource_class.instance_of? Resolv::DNS::Resource::IN::A进行检查，这是不起作用的。我要怎么检查这个？

Answer 1

你的resource_class是一个类。它不是Resolv::DNS::Resource::IN::A的一个实例；它是Resolv::DNS::Resource::IN::A。因此，简单的比较就足够了：

resource_class == Resolv::DNS::Resource::IN::A