javascript - 如何允许任何泛型作为参数传递？

Question

如何声明一个允许传递任何泛型类型的类型？这是我正在尝试做的事情的简化示例:

/* @flow */

type InterfaceType = {
    var1 : number,
};

type ActualType = InterfaceType & {
    var2 : string,
};

class StateOne<T : InterfaceType> {
    constructor(arg : T) : StateOne<T> {
        return this;
    }
}

class StateTwo {
    constructor(stateOne : StateOne<InterfaceType>) : StateTwo {
        return this;
    }
}

let variable : ActualType = {
    var1: 1,
    var2: "text",
}

let stateOne : StateOne<ActualType> = new StateOne(variable);
let stateTwo : StateTwo = new StateTwo(stateOne);

这是我收到的流程错误:

    29: let stateTwo : StateTwo = new StateTwo(stateOne);
                                               ^ Cannot call `StateTwo` with `stateOne` bound to `stateOne` because property `var2` is missing in `InterfaceType` [1] but exists in object type [2] in type argument `T` [3].
        References:
        18:     constructor(stateOne : StateOne<InterfaceType>) : StateTwo {
                                              ^ [1]
        7: type ActualType = InterfaceType & {
                                             ^ [2]
        11: class StateOne<T : InterfaceType> {
                           ^ [3]

我的想法是，自从 ActualType是更具体的InterfaceType ，那么我就可以通过StateOne<ActualType>作为需要 StateOne<InterfaceType> 的参数.

Answer 1

我能够通过向构造函数添加泛型来解决该错误:

class StateTwo {
    constructor<T : InterfaceType>(stateOne : StateOne<T>) : StateTwo {
        return this;
    }
}