Typescript 在使用Typescribe时，是否有一种方法可以推断映射类型的键

我有一个数据 struct 来代表我的业务操作:

const operations = {
    list: {
        request: "a",
        response: "b",
    },
    get: {
        request: "a",
        response: "b",
    },
} as const;

我想创建一个函数，它将接受基于上面operations的回调:

type Ops = typeof operations;

type Callbacks = {
    [Property in keyof Ops]: (
        param: Ops[Property]["request"]
    ) => Ops[Property]["response"];
};

现在，如果我想定义我的回调，如果我错过了其中的任何一个，编译器都会抱怨:

const callbacks: Callbacks = {

};
// ^^^--- Type '{}' is missing the following properties from type 'Callbacks': list, get

现在我的问题是，我想创建另一个类型，以便我可以类型判断operations对象的 struct :

interface OperationDescriptor<A, B> {
    request: A;
    response: B;
}

type Operations = {
    [key: string]: OperationDescriptor<any, any>;
};

const operations: Operations = {
    list: {
        request: "a",
        response: "b",
    },
    get: {
        request: "a",
        response: "b",
    },
} as const; // okay

const badOperations: Operations = {
    list: {
        request: "a",
        response: "b",
    },
    get: { // error, missing response prop
        request: "a",        
    },
} as const; 

但当我这样做时，编译器将不再抱怨，因为Operations不知道我在operations中的键.有没有办法既有我的蛋糕又吃它，例如:

• 定义一个类型，我可以用它来判断operations和
• 有一个Callbacks类型会根据operations的 struct 对我的回调函数进行类型判断？