在 TypeScript 中是否有相当于“密封”或“最终”的东西？
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Is there an equivalent to "sealed" or "final" in TypeScript?
提问by bubbleking
I'm trying to implement a method in a super class that should be available for use, but not changeable, in sub classes. Consider this:
我正在尝试在超类中实现一个方法，该方法应该可以在子类中使用，但不可更改。考虑一下：
export abstract class BaseClass {
universalBehavior(): void {
doStuff(); // Do some universal stuff the same way in all sub classes
specializedBehavior(); // Delegate specialized stuff to sub classes
}
protected abstract specializedBehavior(): void;
}
My intention would be that any sub class of BaseClass would not only be free to omit implementation of universalBehavior(), but not even be allowed to provide an implementation. Is this not (yet) possible in TypeScript? Intellisense complains when I omit the implementation in my sub classes. The best I can seem to do is this:
我的意图是 BaseClass 的任何子类不仅可以自由省略 的实现universalBehavior()，而且甚至不允许提供实现。这在 TypeScript 中（还）不可能吗？当我省略子类中的实现时，Intellisense 会抱怨。我似乎能做的最好的是：
export class SubClass extends BaseClass {
universalBehavior(): void {
super.universalBehavior();
}
specializedBehavior(): void {
// sub class' implementation
}
}
Obviously this is problematic because I have to ensure that no sub class ever implements universalBehavior()with anything other than a call to super.universalBehavior().
显然这是有问题的，因为我必须确保没有子类universalBehavior()除了调用super.universalBehavior().
采纳答案by bubbleking
No, at the time of this writing there is not. There is a proposal for such a keyword which is still being considered, but may or may not ever be implemented.
不，在撰写本文时还没有。有一个关于这样一个关键字的提议仍在考虑中，但可能会或可能永远不会实施。
See:
看：
回答by Asher Garland
Example of implementation hack of 'sealed method' as readonly property of type function which throws compiler error when attempting to override in extended class:
将“密封方法”作为只读属性的“密封方法”的实现示例，当尝试在扩展类中覆盖时会引发编译器错误：
abstract class BaseClass {
protected element: JQuery<HTMLElement>;
constructor(element: JQuery<HTMLElement>) {
this.element = element;
}
readonly public dispose = (): void => {
this.element.remove();
}
}
class MyClass extends BaseClass {
constructor(element: JQuery<HTMLElement>) {
super(element);
}
public dispose(): void { } // Compiler error: "Property 'dispose' in type 'MyClass' is not assignable to the same property in base type 'BaseClass'"
}
TypeScript 2.0 supports "final" classes through using of private constructor:
TypeScript 2.0 通过使用私有构造函数来支持“final”类：
class A {
private constructor(){}
}
class B extends A{} //Cannot extend a class 'A'. Class constructor is marked as private.ts(2675)
回答by Qwerty Pnk
// I use this workaround:
export default class CreateHandler extends BaseHandler {
// final prop used as method
public readonly create = (blueprint: Blueprint): Response<Record> => {
return blueprint.create();
};
// normal method
public store(blueprint: Blueprint): Response<Record> {
return this.response(blueprint.create());
}
}
