TypeScript requires that when explicitly specifying the return type of a generator, the `Generator.prototype.return` function must accept

I am utilizing a generator function to serve as a consumer for accumulating strings:

function *concatStrings(): Generator<void, string, string> {
  let result = '';
  try {
    while (true) {
      const data = yield;
      result += data;
    }
  } finally {
    return result;
  }
}

const g = concatStrings();
g.next();
g.next('abc');
const r = g.return();
console.log(r);

Everything works smoothly, however TypeScript raises an error

Expected 1 arguments, but got 0.ts(2554)
lib.es2015.generator.d.ts(26, 12): An argument for 'value' was not provided.

Playground example

How can I modify the code so that TypeScript does not raise any errors, while still maintaining the correct types?

Answer №1

The issue arises from the fact that you are "overriding" the typical process of the `return` method by utilizing a `return` statement within a `finally` block, which discards the provided value and instead returns `result`. This inconsistency leads to a mismatch in signatures. In the usual scenario, the parameter value of the `return` method corresponds to the `TReturn` of the generator (example). As per MDN:

Generator.prototype.return()

Functions like a `return` statement inserted in the generator's body at its current suspended position, terminating the generator and enabling it to execute any necessary cleanup operations when used with a `try`...`finally` block.

(Cleanup tasks typically do not involve a `return` statement.)

If it helps, I suggest defining and using your generator slightly differently — simply utilize the result of `next`, without requiring that final `return`, as demonstrated below:

function *concatStrings(): Generator<string, void, string> {
// −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−^^^
    let result = '';
    // (Note the absence of `try`/`finally` since no clean up resources exist)
    while (true) {
        const data = yield result;
// −−−−−−−−−−−−−−−−−−−−−−−^^^^^^
        result += data;
    }
}

const g = concatStrings();
let r: IteratorResult<string, void>;
g.next();
r = g.next("a");
console.log(r.value); // "a"
r = g.next("b");
console.log(r.value); // "ab"
r = g.next("c");
console.log(r.value); // "abc"

Playground example

This approach also resolves the typing issue by adhering more closely to conventional practices. :-)

Answer №2

The function known as return requires a parameter for its execution. In cases where the parameter type is designated as void, it is not mandatory to provide a value. An alternative approach is defining the return type as string | void. This configuration allows for flexibility in specifying the return type:

function *combineStrings(): Generator<void, string | void, string> {
  let result = '';
  try {
    while (true) {
      console.log("Loop")
      const data = yield;
      result += data;
    }
  } finally {
    console.log("Cleaning up")
    return result;
  }
}

const strings = combineStrings();
strings.next();
strings.next('abc');
const output = strings.return();  
if(typeof output.value ===  "string")  {
  console.log(output.value);
}

Playground Link

In this case, output.value may be of type void | string, which is not ideal. Consequently, you must verify if it is specifically of type string before utilizing it.

An alternative solution involves passing null! as an argument to facilitate the return statement: strings.return(null!); (Playground Link)

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