TypeScript中如何同时校验操作对象结构与回调类型？

我有一个表示业务操作的数据结构：

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

我希望创建一个可接受基于上述operations的回调的类型：

type Ops = typeof operations;

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

此时若定义回调时遗漏任何项，编译器会报错：

const callbacks: Callbacks = {

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

现在我想创建另一个类型来校验operations对象的结构：

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; // 正常

const badOperations: Operations = {
    list: {
        request: "a",
        response: "b",
    },
    get: { // 报错，缺少response属性
        request: "a",        
    },
} as const; 

但这样做后，编译器不再对回调遗漏项报错，因为Operations类型不知道operations的具体键。请问是否有办法兼顾以下两点：

1. 定义类型以校验operations对象的结构；
2. 拥有能基于operations结构校验回调函数的Callbacks类型？

可以通过泛型工具函数实现两者兼顾，既校验operations的结构符合OperationDescriptor规范，又保留其具体的键和字面量类型信息。

步骤1：定义结构校验的泛型函数 #

创建一个泛型函数，约束传入的对象必须符合{ [key: string]: OperationDescriptor<any, any> }结构，同时保留原对象的具体类型：

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

// 泛型函数：校验结构的同时保留原对象的具体类型
const createOperations = <T extends { [K in keyof T]: OperationDescriptor<any, any> }>(ops: T) => ops;

步骤2：用该函数创建operations对象 #

这样生成的operations既会被校验结构合法性，又会完整保留所有具体的键和字面量类型：

const operations = createOperations({
    list: {
        request: "a",
        response: "b",
    },
    get: {
        request: "a",
        response: "b",
    },
} as const); // 结构合法，正常通过

// 结构错误时直接报错
const badOperations = createOperations({
    list: {
        request: "a",
        response: "b",
    },
    get: {
        request: "a",
        // 此处报错：Property 'response' is missing in type '{ request: "a"; }' but required in type 'OperationDescriptor<any, any>'
    },
} as const);

步骤3：基于operations生成Callbacks类型 #

此时typeof operations仍然保留了具体的键和类型信息，之前的Callbacks类型可以正常工作，遗漏回调项时编译器会报错：

type Ops = typeof operations;

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

// 遗漏项会报错：Type '{}' is missing the following properties from type 'Callbacks': list, get
const callbacks: Callbacks = {

};

原理说明 #

泛型函数createOperations的泛型参数T extends { [K in keyof T]: OperationDescriptor<any, any> }，会约束传入的ops必须满足每个属性都是OperationDescriptor类型；同时TypeScript会自动推导T为传入对象的具体字面量类型，不会丢失键和类型信息，从而同时实现了结构校验和保留具体类型的需求。

内容的提问来源于stack exchange，提问作者Adam A