typescript - 如何从 typescript 中的类型中排除 getter only 属性

Question

类中的 getter 是只读属性，因此从以下代码中抛出类型错误是有意义的。

class Car {
    engine: number;
    get hp() {
        return this.engine / 2;
    }
    get kw() {
        return this.engine * 2;
    }
}

function applySnapshot(
    car: Car,
    snapshoot: Partial<Car> // <-- how to exclude readonly properties?
) {
    for (const key in snapshoot) {
        if (!snapshoot.hasOwnProperty(key)) continue;
        car[key as keyof Car] = snapshoot[key as keyof Car];
        // Cannot assign to 'hp' because it is a constant or a read-only property.
    }
}

有没有办法将只写属性转换为类型并排除所有 getter？

example in playground

Answer 1

虽然 readonly 不直接影响类型是否可赋值，但会影响它们是否相同。为了测试两种类型是否相同，我们可以滥用 (1) 条件类型的可分配性规则，这要求 extends 之后的类型相同，或者 (2) 交集类型的推理过程，从双方抛出相同的类型。然后我们只需像 Titian Cernicova-Dragomir 的回答中那样使用映射类型依次查看 Car 的每个属性，看看它是否与自身的可变版本相同。

// https://github.com/Microsoft/TypeScript/issues/27024#issuecomment-421529650
type IfEquals<X, Y, A, B> =
    (<T>() => T extends X ? 1 : 2) extends
    (<T>() => T extends Y ? 1 : 2) ? A : B;

// Alternatively:
/*
type IfEquals<X, Y, A, B> =
    [2] & [0, 1, X] extends [2] & [0, 1, Y] & [0, infer W, unknown]
    ? W extends 1 ? B : A
    : B;
*/

type WritableKeysOf<T> = {
    [P in keyof T]: IfEquals<{ [Q in P]: T[P] }, { -readonly [Q in P]: T[P] }, P, never>
}[keyof T];
type WritablePart<T> = Pick<T, WritableKeysOf<T>>;

class Car {
    engine: number;
    get hp() {
        return this.engine / 2;
    }
    get kw() {
        return this.engine * 2;
    }
}

function applySnapshot(
    car: Car,
    snapshoot: Partial<WritablePart<Car>>
) {
    let key: keyof typeof snapshoot;
    for (key in snapshoot) {
        if (!snapshoot.hasOwnProperty(key)) continue;
        car[key] = snapshoot[key];
    }
}