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C#:获取参数所属的类型

[英]C#: Get the Type that the parameter belongs to

public abstract class Vehicle
{
    protected void SomeMethod<T>(String paramName, ref T myParam, T val)
    {
        //Get the Type that myParam belongs to...
        //(Which happens to be Car or Plane in this instance)
        Type t = typeof(...);
    }
}

public class Car : Vehicle
{
    private String _model;
    public String Model
    {
        get { return _model; }
        set { SomeMethod<String>("Model", ref _model, value); }
    }
}

public class Plane: Vehicle
{
    private Int32 _engines;
    public In32 Engines
    {
        get { return _engines; }
        set { SomeMethod<Int32>("Engines", ref _engines, value); }
    }
}

Is it possible to do what I'm looking for... that is, get t be typeof(Car) or typeof(Plane) using the referenced parameter myParam somehow? 是否可以做我要寻找的...就是以某种方式使用引用的参数myParam获得typeof(Car)或typeof(Plane)?

Oh, and I would like to avoid having to pass in a 'this' instance to SomeMethod or adding another Generic constraint parameter if I can. 哦,如果可以的话,我想避免将'this'实例传递给SomeMethod或添加另一个Generic约束参数。

You don't need to pass in this - it's already an instance method. 您不需要传递this -它已经是一个实例方法。

Just use: 只需使用:

Type t = this.GetType();

That will give the actual type of vehicle, not Vehicle . 这将给出车辆的实际类型,而不是Vehicle

You can call GetType() which will operate on the current instance. 您可以调用将在当前实例上运行的GetType() It's a little misleading because the code lives in the base class, but it will correctly get the inherited class type for you at runtime. 这有点误导,因为代码位于基类中,但是它将在运行时为您正确获取继承的类类型。

Type t = this.GetType(); 
/*or equivlently*/
Type t = GetType();

On a side note you don't have to pass the type into SomeMethod it will be infered for you by the compiler. 另外,您不必将类型传递给SomeMethod ,编译器会为您SomeMethod该类型。

public class Car : Vehicle 
{ 
    private String _model; 
    public String Model 
    { 
        get { return _model; } 
        set { SomeMethod("Model", ref _model, value); } 
    } 
} 

使SomeMethod为非泛型,然后Type t = this.GetType()

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