[英]Access a object's class's static methods in Typescript?
class Base {
static f(){console.log('Base')}
}
class A extends Base {
static f(){console.log('A')}
}
class B extends Base {
static f(){console.log('B')}
}
let obj: A|B = new A()
obj.<what to put here>.f()
I don't know the exact class for obj, I need to print A or just call f()
for correct class of obj.我不知道 obj 的确切 class,我需要打印 A 或只调用
f()
以获得正确的 obj 的 class。
I don't need just the class name, that is for example.例如,我不需要 class 名称。 I am doing more complex things.
我正在做更复杂的事情。
prototype, typeof, constructor
all seem to be syntax errors. prototype, typeof, constructor
似乎都是语法错误。
Both Object.getPrototypeOf() (replacement to the now deprecated Object.prototype.__proto__
) or Object.prototype.constructor
should work: Object.getPrototypeOf() (替换为现已弃用的
Object.prototype.__proto__
)或Object.prototype.constructor
.
Object.getPrototypeOf(obj).constructor.f();
obj.constructor.f();
Actually:实际上:
Object.getPrototypeOf(obj).constructor === obj.constructor; // true
Here you can see the compiled source in action:在这里,您可以看到编译后的源代码:
class Base { static f() { console.log('Base'); } } class A extends Base { static f() { console.log('A'); } } class B extends Base { static f() { console.log('B'); } } const objBase = new Base(); const objA = new A(); const objB = new B(); Object.getPrototypeOf(objBase).constructor.f(); objBase.constructor.f(); Object.getPrototypeOf(objA).constructor.f(); objA.constructor.f(); Object.getPrototypeOf(objB).constructor.f(); objB.constructor.f(); console.log(Object.getPrototypeOf(objB).constructor === objB.constructor); console.log(Object.getPrototypeOf(objB) === B.prototype);
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Note static properties exist in classes but not in instances.注意 static 属性存在于类中,但不存在于实例中。
Therefore, if you want to go from obj
to its prototype, you should call Object.getPrototypeOf(obj)
, not obj.prototype
, which is a completely different thing.因此,如果你想 go 从
obj
到它的原型,你应该调用Object.getPrototypeOf(obj)
,而不是obj.prototype
,这是完全不同的事情。
The .prototype
property only exists in functions and, when instantiating a new object using new
and a call to that constructor function ( new A()
), will become the newly created object's prototype (the deprecated .__proto__
). .prototype
属性仅存在于函数中,并且当使用new
实例化新的 object 并调用该构造函数 function ( new A()
) 时,将成为新创建的对象的原型(已弃用的.__proto__
)。
In your example:在您的示例中:
obj.prototype; // undefined
A; // class A { static f() { ... } }
A.protoype; // { constructor: class A { ... } }
A.protoype.constructor; // class A { static f() { ... } }
A.protoype.constructor === A; // true
obj.constructor; // class A { static f() { ... } }
obj.constructor === A; // true
Object.getPrototypeOf(obj) === A.prototype; // true
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