typescript - 展开运算符类型不安全

Question

我有一个函数应该返回某种类型，但使用扩展运算符导致它分配一个拼写错误的键。

interface State {
    fieldA: number,
    fieldB: string
}

const reducer: (state: State, action: {payload: string}) => State = (state, action) => {

    // please note that those variables are not desired
    const tmp = { ...state, feildB: action.payload }; // No compile time error, :(

    // This is too verbose... but works
    const tmp2 = Object.assign<State, Partial<State>>(state, {feildB: action.payload}) // ERROR - this is what I need
    return tmp
}
const t = reducer({fieldA: 1, fieldB: 'OK'}, {payload: 'Misspelled'}) // Misspelled
console.log("feildB", (t as any).feildB) // Misspelled
console.log("fieldB", (t as any).fieldB) // OK

有没有办法使其类型安全，同时将样板代码保持在最低限度？

Playground code HERE

Answer 1

TypeScript 正在做它应该做的事情。在您的例子中，您正在创建一个新对象 tmp，其新类型具有 3 个字段，即:

interface State {
    fieldA: number;
    fieldB: string;
}
interface Tmp {
    fieldA: string;
    fieldB: string;
    payload: string;
}

换句话说，传播运算符执行以下操作:

interface Obj {
    [key: string]: any;
}

const spread = (...objects: Obj[]) => {
    const merged: Obj = {};

    objects.forEach(obj => {
        Object.keys(obj).forEach(k => merged[k] = obj[k]);
    });

    return merged;
}

传播运算符正在为您创建一种新类型的对象；如果你想推断类型，那么你应该这样做:

// this now throws an error
const tmp: State = { ...state, feildB: action.payload };