Question

我想将类引用传递给将创建并返回该类实例的函数。但是，我想根据它实现的接口类型检查传入的类。

accepted answer to 'How does typescript interfaces with construct signatures work?'描述了一种依赖于构造函数签名的结构类型检查的方法。当然，这个问题特定于构造函数签名，所以也许我正在咆哮错误的树，因为在我的例子中：

实现IHappyController的类不共享相同的构造函数签名。

当不相关接口中的构造函数匹配时，很容易产生误报。

interface IHappyController
{
  name:string;
}

interface IHappyControllerConstructable
{
  // This constructor is compatible with JoyfulController
  new ( name:String, joyfulSong:string, joyfulColor:string ):IHappyController;

  // This constructor is compatible with DelightfulController
  //new ( name:String, time:Date ):IHappyController;
}

interface ISadController
{
  name:string;
  reason:string;
}

interface ISadControllerConstructable
{
  // This constructor is compatible with DismalController, signature is same as DelightfulController
  new ( name:String, time:Date ):ISadController;
}

class JoyfulController implements IHappyController
{
  constructor( public name:string, joyfulSong:string, joyfulColor:string )
  {

  }
}

class DelightfulController implements IHappyController
{
  constructor( public name:string, time:Date )
  {

  }
}

class DismalController implements ISadController
{
  reason:string;

  constructor( public name:string, time:Date )
  {

  }
}

// happyMaker should return an instance of the passed class, but only if the instance implements IHappyController
var happyMaker = ( clazz:IHappyControllerConstructable ):IHappyController => {
  var instance:IHappyController = ServiceLocator.getByClass( clazz );
  return instance;
}

happyMaker(JoyfulController);       // Expect Success, it implements IHappyController
happyMaker(DelightfulController);   // Expect Success, it implements IHappyController
happyMaker(DismalController);       // Expect Error, it implements ISadController

更简单的解决方案？

我猜（希望）有一个更简单的解决方案，根本不依赖于I*Constructable接口。但是happyMaker()的方法签名应该是什么？

interface IHappyController
{
  name:string;
}

interface ISadController
{
  name:string;
  reason:string;
}

class JoyfulController implements IHappyController
{
  constructor( public name:string, joyfulSong:string, joyfulColor:string )
  {

  }
}

class DelightfulController implements IHappyController
{
  constructor( public name:string, time:Date )
  {

  }
}

class DismalController implements ISadController
{
  reason:string;

  constructor( public name:string, time:Date )
  {

  }
}

// happyMaker should return an instance of the passed class, but only if the instance implements IHappyController
var happyMaker = ( clazz:/* WHAT GOES HERE? */ ):IHappyController => {
  var instance:IHappyController = ServiceLocator.getByClass( clazz );
  return instance;
}

happyMaker(JoyfulController);       // Expect Success, it implements IHappyController
happyMaker(DelightfulController);   // Expect Success, it implements IHappyController
happyMaker(DismalController);       // Expect Error, it implements ISadController

Answer 1

所写的示例无效，因为ISadController是IHappyController的子类型而TypeScript使用结构类型，因此对IHappyController有效的任何内容都将有效与ISadController。

有关

如果您调整IHappyController以使其不是ISadController的超类型，则可以执行此操作：

interface IHappyController {
  happyName: string;
}

interface ISadController {
  name:string;
  reason:string;
}

class JoyfulController implements IHappyController {
  constructor(public happyName:string, joyfulSong:string, joyfulColor:string) { }
}

class DelightfulController implements IHappyController {
  constructor(public happyName:string, time:Date) { }
}

class DismalController implements ISadController {
  reason:string;
  constructor(public name:string, time:Date) { }
}

declare var ServiceLocator;
var happyMaker = <T extends IHappyController>( clazz: { new(...args: any[]): T } ): T => {
  var instance = ServiceLocator.getByClass( clazz );
  return <T>instance;
}

happyMaker(JoyfulController); // OK
happyMaker(DelightfulController); // OK
happyMaker(DismalController); // Error

根据接口键入check class reference

更简单的解决方案？

1 个答案: