Question

我想做一个抽象类字典来存储扩展抽象类的类，而不是实例。但我有一个问题。

myStateComputation :: IState Int String Bool
myStateComputation =
    -- poor man's do notation. You could use RebindableSyntax
    get              >>>= \s ->
    put (show s)     >>>
    ireturn (s > 5)

main = print $ runIState myStateComputation 3
-- ("3", False)

那么，您认为我怎么能得到这个？

我不想从AbstractClass中删除abstract修饰符，因为我需要它。

Answer 1

abstract class  AbstractClass{
   foo(){
      // do someThing
   }
 }

class ConcreteClass extends AbstractClass{
    foo(){
        super.foo()
    }
}


interface ClassDicctionary {
    [index:string]: new() => AbstractClass; // <---
}


let dicctionary : ClassDicctionary = { "ONE" :  ConcreteClass }

new dicctionary["One"]().foo();

Answer 2

abstract class  AbstractClass{



   constructor( foo:string){

   } 

   foo(){
      // do someThing
   }
 }

class ConcreteClass extends AbstractClass{


   constructor( foo:string, bar: number){
       super(foo)
   } 

    foo(){
        super.foo()
    }
}


interface ClassDicctionary {
    [index:string]: new( ...args) => AbstractClass; // <---
}


let dicctionary : ClassDicctionary = { "ONE" :  ConcreteClass }

new dicctionary["One"]().foo(); // No error but It needs some arguments

TypeScript中的抽象类字典

2 个答案: