typescript - 是否可以让函数接受类实例，但不接受普通对象？

Question

我想编写一个函数，它将接受类实例作为参数，但不允许普通的、匿名类型的对象。

例如:

class Dog {
    constructor(name: string, breed: "husky" | "boxer") {
        this.name = name;
        this.breed = breed;
    }
    name: string;
    breed: "husky" | "boxer";
}

class Cat {
    constructor(name: string, breed: "siamese" | "persian") {
        this.name = name;
        this.breed = breed;
    }
    name: string;
    breed: "siamese" | "persian";
}

function pat(pet: NoPlainObjects) {
    document.write(pet.constructor.name);
}

pat(new Dog('Fido', 'boxer')); //works
pat(new Cat('Spot', 'persian')); //works

pat({name: 'snuffaluffagus'}); //compile error

Answer 1

Is it possible to have a function accept class instances, but not accept plain objects?

目前这是不可能的，因为 TypeScript 使用结构子类型。这意味着使用类构造函数创建的对象与普通的旧 JavaScript 对象文字兼容，前提是两者都具有兼容的属性。

这是文档所说的about type compatibility :

The basic rule for TypeScript’s structural type system is that x is compatible with y if y has at least the same members as x. To check whether y can be assigned to x, the compiler checks each property of x to find a corresponding compatible property in y... The same rule for assignment is used when checking function call arguments. [empahsis added]

您的问题提出了一个问题:为什么您希望函数接受类实例并拒绝普通的旧 JavaScript 对象？