[英]Interface and abstract c# inheritance
I am encountering a problem with interfaces. 我遇到接口问题。 I wish to chain a method which derives its chainable method from an abstract class, which implements an interface. 我希望链接一个方法,该方法从实现接口的抽象类派生其可链接方法。
public interface IBaseInterface {
public IBaseInterface ChainableMethod()
}
public abstract AbstractClassThatHelps<T> where T: IBaseInterface n {
public T ChainableMethod() {
return (T) this;
}
}
public interface IDerived : IBaseInterface { }
public class DerivedClass : AbstractClassThatHelps<IDerived>, IDerived { }
IDerived derived = new DerivedClass();
derived.ChainableMethod().ChainableMethod();
The problem I face here: why can't T
be returned when it is shown to implement the contract IModel? 我在这里面临的问题:为什么在显示实现合同IModel时不能返回T
?
How would I solve this differently? 我该如何解决呢? I wish to have type safety but I am forced for all derived classes to return IBaseInterface instead of their own interface. 我希望具有类型安全性,但是我被迫让所有派生类返回IBaseInterface而不是它们自己的接口。
Actual implementation: We have multiple models ( DerivedClass
's) which implement their respective IDerived
for dependency injection. 实际实现:我们有多个模型( DerivedClass
),它们实现了各自的IDerived
用于依赖项注入。 These need helpers because i dont want to repeat myself.. So we use AbstractClassThatHelps
as a base, but because we are dealing with chainable methods we need this base class to know what to return, so therefore generics. 这些需要帮助者,因为我不想重复自己。.因此,我们使用AbstractClassThatHelps
作为基础,但是由于我们正在处理可链接的方法,因此我们需要该基础类来知道要返回的内容,因此要知道泛型。 IBaseInterface
can be seen as IModel
. IBaseInterface
可以看作IModel
。 Where ChainableMethod
can be seen as GetAll()
for example. 例如,其中ChainableMethod
可以视为GetAll()
。
In order for the following code to work AbstractClassThatHelps<T>
must implement IBaseInterface
. 为了使以下代码起作用, AbstractClassThatHelps<T>
必须实现IBaseInterface
。 How can you return this
, if this
is not IBaseInterface
你怎么能回this
,如果this
不是IBaseInterface
public abstract AbstractClassThatHelps<T> where T: IBaseInterface n{
public T ChainableMethod(){
return this;
}
}
Edit: I am not user what this design solves but here is my attempt at what you are trying to achieve→ 编辑:我不是用户此设计解决的问题,但这是我对您要实现的目标的尝试→
public interface IBaseInterface
{
IBaseInterface ChainableMethod();
}
public abstract class AbstractClassThatHelps<T>:IBaseInterface where T : IBaseInterface{
public T ChainableMethod()
{
IBaseInterface i = this;
return (T)i.ChainableMethod();
}
IBaseInterface IBaseInterface.ChainableMethod()
{
return this;
}
}
public class Concrete : AbstractClassThatHelps<Concrete>
{
}
您可以返回T
的实例 ,但是返回类型不能为T
而必须为IBaseInterface
因为这是接口所需要的。
This is working, your code was full of syntax errors: 这可以正常工作,您的代码充满语法错误:
public interface IBaseInterface
{
IBaseInterface ChainableMethod();
}
public abstract class AbstractClassThatHelps : IBaseInterface
{
public IBaseInterface ChainableMethod()
{
return this;
}
}
public interface IDerived : IBaseInterface
{
}
public class DerivedClass : AbstractClassThatHelps, IDerived
{
}
internal static class Program
{
public static void Main()
{
IDerived derived = new DerivedClass();
derived.ChainableMethod().ChainableMethod();
}
}
you could also try this: 您也可以尝试以下方法:
public interface IBaseInterface
{
IBaseInterface ChainableMethod();
}
public abstract class AbstractClassThatHelps<T> : IBaseInterface where T : class, IBaseInterface
{
public T ChainableMethod()
{
return this as T;
}
IBaseInterface IBaseInterface.ChainableMethod()
{
return ChainableMethod();
}
}
public interface IDerived : IBaseInterface
{
IDerived Hello();
}
public class DerivedClass : AbstractClassThatHelps<IDerived>, IDerived
{
public IDerived Hello()
{
Console.WriteLine("Hello");
return this;
}
}
internal static class Program
{
public static void Main()
{
AbstractClassThatHelps<IDerived> derived = new DerivedClass();
derived.ChainableMethod().Hello().ChainableMethod();
}
}
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