将对象转换为泛型类型 [英] Cast object to a generic type

问题描述

我有一个或多或少的泛型类：

我有一段时间没有睡，所以这可能比我想的要容易。 p>

  public class Reference< T>其中T：APIResource //<  -  APIResource是抽象的btw 
 {
 private T _value = null; 
 public T value 
 {
 get {return _value; } 
} 
} 
  

另外，在一个自定义序列化方法中，传入一个对象，它实际上是一个参考<（something）> 的实例。我只是想跳到每个 Reference<> 对象具有的value属性，所以我想去：

  string serialize（object o）
 {
 return base.serialize（（（Reference<>）o）.value）; 
 
  

当然，生活并不那么简单，因为编译器会说： b
$ b

使用泛型类型Reference< T>需要1个类型参数

我该怎么做我想做的事？

  interface IReference< out> 
 
 T> T：ApiResource {
 T Value {get; } 
} 
  

然后您可以投射 IReference< Anything> 到 IReference< object> 或 IReference< ApiResource> 。

I haven't slept in a while so this is probably easier than I think it is.

I have a generic class that's more or less this:

public class Reference<T> where T : APIResource //<- APIResource is abstract btw
{
    private T _value = null;
    public T value
    { 
        get { return _value; }
    }
}

Elsewhere, in a custom serialize method, someone is passing in an object that is actually an instance of Reference<(something)>. I simply want to skip to the "value" property that every Reference<> object has, so I want to go:

string serialize(object o)
{
    return base.serialize( ((Reference<>) o).value );
}

Of course, life isn't that simple because as the compiler puts it:

using the generic type "Reference<T>" requires 1 type arguments

How can I do what I want to do?

解决方案

You can create a covariant generic interface with the property:

interface IReference<out T> where T : ApiResource {
    T Value { get; }
}

You can then cast IReference<Anything> to IReference<object> or IReference<ApiResource>.

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