我有一个存储的序列化值和类型的类。我想有一个属性/方法返回已浇铸的值：

I have a class that stores a serialized value and a type. I want to have a property/method returning the value already casted:

public String Value { get; set; }

public Type TheType { get; set; }

public typeof(TheType) CastedValue { get { return Convert.ChangeType(Value, typeof(_Type)); }

这是可能在C＃？

Is this possible in C#?

推荐答案

这是可能的，如果包含该属性的类是通用的，您声明使用泛型参数属性：

It's possible if the class containing the property is generic, and you declare the property using the generic parameter:

class Foo<TValue> {
    public string Value { get; set; }
    public TValue TypedValue {
        get {
            return (TValue)Convert.ChangeType(Value, tyepof(TValue));
        }
    }
}

另一种方法是使用一个通用的方法来代替：

An alternative would be to use a generic method instead:

class Foo {
    public string Value { get; set; }
    public Type TheType { get; set; }

    public T CastValue<T>() {
         return (T)Convert.ChangeType(Value, typeof(T));
    }
}

您也可以使用 System.ComponentModel.TypeConverter 类进行转换，因为它们允许一个类来定义它自己的转换器。

You can also use the System.ComponentModel.TypeConverter classes to convert, since they allow a class to define it's own converter.

修改：请注意，调用泛型方法时，必须指定泛型类型参数，因为编译器无法推断它：

Edit: note that when calling the generic method, you must specify the generic type parameter, since the compiler has no way to infer it:

Foo foo = new Foo();
foo.Value = "100";
foo.Type = typeof(int);

int c = foo.CastValue<int>();

您必须知道在编译时的类型。如果你不知道在编译时的类型，那么你必须将其存储在一个对象，在这种情况下，你可以添加以下属性到富类：

You have to know the type at compile time. If you don't know the type at compile time then you must be storing it in an object, in which case you can add the following property to the Foo class:

public object ConvertedValue {
    get {
        return Convert.ChangeType(Value, Type);
    }
}