Javascript 为返回包装函数的函数的vanilla函数添加d.ts
在我的vanilla JS代码中有一个复杂的函数(参见后面的简单示例),它获取Javascript 为返回包装函数的函数的vanilla函数添加d.ts,javascript,typescript,.d.ts,Javascript,Typescript,.d.ts,在我的vanilla JS代码中有一个复杂的函数(参见后面的简单示例),它获取exportd,我想向它添加一个d.ts 到目前为止,我编写.d.ts的尝试还没有将类型从中间阶段转发到最后阶段 函数看起来是这样的(我在保持相同结构的同时尽可能简化了它) 示例用法 const A = factory(1, 2); const fn = (a1, a2) => a1 + ' world' + a2; // (a1: string, a2: string) => string const B
export
d,我想向它添加一个d.ts
到目前为止,我编写.d.ts
的尝试还没有将类型从中间阶段转发到最后阶段
函数看起来是这样的(我在保持相同结构的同时尽可能简化了它)
示例用法
const A = factory(1, 2);
const fn = (a1, a2) => a1 + ' world' + a2; // (a1: string, a2: string) => string
const B = A(fn); // (i: string, j: string) => Promise<string>
B('hello', '!'); // warns 3 and returns a Promise<string> "hello world!"
但是,
- 当查看
时,关于B
的所有类型信息都丢失了,而当我期望fn
(字符串,字符串)=>承诺时,我只剩下
任何[]=>承诺
- 返回类型的描述重复多次
declare function factory(x: number): wrapper;
declare function factory(x: number, y: number): wrapper;
interface wrapper {
(fn: function): wrapped;
}
interface wrapped {
(...args: any): Promise<any>;
}
声明函数工厂(x:number):包装器;
声明函数工厂(x:number,y:number):包装器;
接口包装器{
(fn:功能):包装;
}
界面包裹{
(…args:any):承诺;
}
如果我正确理解了这个问题,您希望最终函数的参数与作为参数传递的函数的参数相同。虽然不能对任意数量的参数执行此操作,但您可以为最多包含4个参数的函数定义重载(例如),并在以后需要时添加更多参数:
declare function factory(x: number): Promiseify;
declare function factory(x: number, y: number): Promiseify;
interface Promiseify {
<T>(fn: () => T): () => Promise<T>;
<T, U1>(fn: (arg1: U1) => T): (arg1: U1) => Promise<T>;
<T, U1, U2>(fn: (arg1: U1, arg2: U2) => T): (arg1: U1, arg2: U2) => Promise<T>;
<T, U1, U2, U3>(fn: (arg1: U1, arg2: U2, arg3: U3) => T): (arg1: U1, arg2: U2, arg3: U3) => Promise<T>;
<T, U1, U2, U3, U4>(fn: (arg1: U1, arg2: U2, arg3: U3, arg4: U4) => T): (arg1: U1, arg2: U2, arg3: U3, arg4: U4) => Promise<T>;
}
const A = factory(1, 2);
const fn = (a1: string, a2: number) => a1 + ' world' + a2; // (a1: string, a2: number) => string
const B = A(fn); // (arg1: string, arg2: number) => Promise<string>
B('hello', 1); // ok
B('hello', '1'); // error
申报功能工厂(x:编号):承诺;
申报功能工厂(x:编号,y:编号):承诺;
接口承诺{
(fn:()=>T:()=>承诺;
(fn:(arg1:U1)=>T:(arg1:U1)=>Promise;
(fn:(arg1:U1,arg2:U2)=>T:(arg1:U1,arg2:U2)=>Promise;
(fn:(arg1:U1,arg2:U2,arg3:U3)=>T:(arg1:U1,arg2:U2,arg3:U3)=>Promise;
(fn:(arg1:U1,arg2:U2,arg3:U3,arg4:U4)=>T:(arg1:U1,arg2:U2,arg3:U3,arg4:U4)=>Promise;
}
常数A=工厂(1,2);
常量fn=(a1:string,a2:number)=>a1+“世界”+a2;//(a1:字符串,a2:数字)=>字符串
常数B=A(fn);//(arg1:string,arg2:number)=>Promise
(你好,1);//好啊
B('hello','1');//错误
如果我理解正确,您希望结果函数的参数与作为参数传递的函数的参数相同?遗憾的是,您不能任意执行,但对于合理的已知长度,这看起来效果很好,谢谢!
declare function factory(x: number): Promiseify;
declare function factory(x: number, y: number): Promiseify;
interface Promiseify {
<T>(fn: () => T): () => Promise<T>;
<T, U1>(fn: (arg1: U1) => T): (arg1: U1) => Promise<T>;
<T, U1, U2>(fn: (arg1: U1, arg2: U2) => T): (arg1: U1, arg2: U2) => Promise<T>;
<T, U1, U2, U3>(fn: (arg1: U1, arg2: U2, arg3: U3) => T): (arg1: U1, arg2: U2, arg3: U3) => Promise<T>;
<T, U1, U2, U3, U4>(fn: (arg1: U1, arg2: U2, arg3: U3, arg4: U4) => T): (arg1: U1, arg2: U2, arg3: U3, arg4: U4) => Promise<T>;
}
const A = factory(1, 2);
const fn = (a1: string, a2: number) => a1 + ' world' + a2; // (a1: string, a2: number) => string
const B = A(fn); // (arg1: string, arg2: number) => Promise<string>
B('hello', 1); // ok
B('hello', '1'); // error