如何传递返回某种类型对象的函数数组.换句话说 - 构造函数数组?
我想做的是这样的:
constructor(name: string, systems: Array){ this.system = new systems[0](); }
但我得到一个错误Cannot use 'new' with an expression whose type lacks a call or construct signature.
,据我所知,我应该以某种方式让编译器知道构造函数返回的对象类型,但我不知道如何.
假设RJM的回答,这是我的意思的一个扩展示例:
interface System{ name: string; } interface Component{ blahblha: string; } class Doh implements Component{ blahblha: string; } class Bar implements System{ name: string = 'bar'; } class Blah implements System{ name: string = 'foo'; } class Foo { systems: Array; constructor(bars: Array<()=>System>) { for (let i in bars){ this.systems.push(new bars[i]()); // error: only a void functions can be called with the 'new' keyword. } } } var foo = new Foo([Blah]); // error: Argument of type 'typeof Blah[]' is not assignable to parameter of type '(() => System[]'. Type 'typeof Blah[]' is not assignable to type '() => System'.
我想确保Foo.systems
仅使用实现System接口的对象实例填充.所以我应该有错误做某事,new Foo([Doh]);
但即使在通过时我也会收到错误Blah
如果要实例化类,则必须使用new关键字引用它的构造函数.
function factory(constructors: { new (): System }[]) {
let constructed = constructors.map(c => new c());
}
interface System {
name: string;
}
class SomeClass implements System {
name: string;
constructor() { }
}
class SomeOtherClass implements System {
name: string;
constructor() { }
}
let a = factory([SomeClass, SomeClass, SomeClass, SomeOtherClass]);
该工厂的功能将得到的满足,可以创造一些构造函数列表系统接口.
使用未实现System的类调用该方法将导致您现在收到的错误.
class AnotherClass {}
factory([AnotherClass]); // error
结构类型也可以,因此您不必显式实现接口.
class AnotherClass {
name: string
}
factory([AnotherClass]);
为了使代码更清洁,您可以将为constructors参数描述的接口移动到接口声明.
// an interface that describes classes that create instances that satisfy the System interface
interface CreatesSystem {
new (): System
}
interface System {
name: string;
}
function factory(constructors: CreatesSystem[]) {
let constructed = constructors.map(c => new c());
}