强制实施泛型类型返回函数

Question

为了说明这个例子，假设我想编写一个函数，将日志记录添加到任何返回 Promise 的函数中。在JS中，我会做这样的事情：

const addLogging = (f) => (...args) => (
  f(...args).then(result => {
    console.log('result:', result);
    return result;
  })
)
const test = addLogging(
  (value) => Promise.resolve(value)
)
test('foo') // logs "​​​​​result: foo​​​​​"

现在我想用打字稿强制执行打字。这是我想到的：

const addLogging = <F extends Function>(f: F): F => (
  (
    (...args: any[]) => (
      (f as any)(...args).then((result: any) => {
        console.log('result:', result);
        return result;
      })
    )
  ) as any
);
// Cool! :)
// type of test is (value: string) => Promise<string>
const test = addLogging(
  (value: string) => Promise.resolve(value),
);
// Less Cool :(
// Not valid, how to prevent it with typings?
const test2 = addLogging(
  (value: string) => value, // should return a promise
);

保留了增强功能的键入，这很好。但首先我必须使用很多any，而且我还想强制addLogging的f参数必须是返回Promise的函数。有没有简单的方法可以用打字稿做到这一点？

Answer 1

您可以更具体地说明对F的约束，您可以指定 这是一个接受任意数量的参数并返回Promise<any>

const addLogging = <F extends (...args: any[]) => Promise<any>>(f: F) => ((
  (...args) =>
    f(...args).then((result: any) => {
      console.log('result:', result);
      return result;
    })
) as F);
//Ok
const test = addLogging(
    (value: string) => Promise.resolve(value),
);
//Error
const test2 = addLogging(
    (value: string) => value, // should return a promise
);