正确键入在 TypeScript 中调用另一个生成器 function 的生成器 function

Question

I have a utility function which is a generator, like this:

function* myGenerator(): Generator<Foo, void> {
  ... // yield Foos
}

Now I want to wrap this utility function in a class, like this:

class MyWrapper {
  *generator(): Generator<Foo, void> {
    // call utils.myGenerator();
  }
}

The problem is: If I type MyWrapper.generator() as a generator with the * notation, I cannot simply implement it as return utils.myGenerator() , because a generator function returns void, and I will get a Type Generator<Foo, void> is not assignable to void error.

I can remove the * from MyWrapper.generator() , have the return type be Generator<Foo, void> , and return utils.myGenerator() :

class MyWrapper {
  generator(): Generator<Foo, void> {
    return utils.myGenerator();
  }
}

... but losing * makes it less clear for the caller that this is a generator function (granted, the return type should give it away, but still).

I can also iterate through the "inner" generator and yield everything from it:

class MyWrapper {
  *generator(): Generator<Foo, void> {
    for (const x of utils.myGenerator()) {
      yield x;
    }
  }
}

... but this extra yielding layer seems silly.

Is there an obvious way I'm missing to have a generator function wrap another generator function and still maintain the * notation, such as a "yield everything from" syntax?

Answer 1

You can use yield* to yield all elements of another generator

type Foo = string
function* myGenerator(): Generator<Foo, void> {
    yield "A";
    yield "B";
}


class MyWrapper {
  *generator(): Generator<Foo, void> {
    yield* myGenerator();
  }
}

console.log([...new MyWrapper().generator()]) // ["A", "B"] 

Playground Link