Question

考虑：

class BasicType
{
    public BasicType() { }
    public T Save<T>() where T : BasicType
    {
        BasicType b = DataContext.Save(this); //Returns a BasicType
        return (T)Activator.CreateInstance(typeof(T), b);
    }
}

class DerivedType : BasicType
{
    public DerivedType(BasicType b) { }
}

public static void Main()
{
    DerivedType d = new DerivedType();
    d = d.Save<DerivedType>();
}

这样可行，但每次调用Save时被强制指定类型都是拖拽。是否有某种方法可以更改BasicType.Save方法，以便它始终返回正在调用Save的实例的实际类型（派生或基础）的实例？

Answer 1

在这种情况下不需要泛型。

我认为这应该足够了：

public BasicType Save()
{
     BasicType b = DataContext.Save(this); //Returns a BasicType
     return (BasicType)Activator.CreateInstance(this.GetType(), b);
}

无论如何你应该小心这个，因为继承的类可能没有预期的构造函数。

最好覆盖保存方法，或者至少是特定部分。

Answer 2

您可以更改BasicType的定义，以便在继承时强制提供T的类型。

这样的事情：

class BasicType<T> where T :  BasicType<T>, new()
{
    public BasicType() { }
    public T Save() 
    {
        T b = new T();
        return (T)Activator.CreateInstance(typeof(T), b);
    }
}

class DerivedType : BasicType<DerivedType>
{
    public DerivedType() { }
}

class Program
{
    static void Main(string[] args)
    {
        DerivedType d = new DerivedType();
        d = d.Save();
    }
}

Answer 3

像这样的东西。

    class BasicType
    {
        public BasicType() 
        { 
        }

        protected virtual T Save<T>()
        {
            BasicType b = DataContext.Save(this); //Returns a BasicType

            return (T)Activator.CreateInstance(typeof(T), b);
        }
    }

    class DerivedType : BasicType
    {
        public DerivedType(BasicType b) 
        { 
        }

        public DerivedType Save()
        {
            return base.Save<DerivedType>();
        }
    }

    public static void Main()
    {
        DerivedType d = new DerivedType(new BasicType());
        d = d.Save();
    }

返回继承类的类型

3 个答案: