TypeScript does not accept an Iterator within a for...of loop

Question

TypeScript does not accept an Iterator within a for...of loop

In my browser, the code works fine even without the typings. However, TypeScript is emitting an error stating that gen() does not have a [Symbol.iterator] method, which is required for it to be considered an Iterable. I find this limitation odd because as far as I know, an Iterator should be a valid object to use with a for...of loop.

function *gen(): Iterator<number> {
  yield 1
  yield 2
  yield 3
}

for (const val of gen()) {
  console.log(val)
}

Could you please help me understand what mistake I am making here?

Edit: After removing the return type from the code above and allowing TypeScript to infer it automatically, I got IterableIterator, which satisfied TypeScript. Does this mean I cannot simply use an Iterator in a for...of loop?

typescript iterator generator

Answer 1

Answer №1

In order to properly handle the series of values, it seems like using Iterable<T> instead of Iterator<T> is necessary. The interface Iterable contains the method [Symbol.iterator] that for-of loops call upon, returning an Iterator with a next() method to generate the values.

function* gen(): Iterable<number> {
  yield 1
  yield 2
  yield 3
}

Currently, TypeScript's handling of Generators appears to be incomplete. While there is a built-in type definition for Generator, it lacks generics (which means the produced values default to any). This would result in an error as it is identified only as an Iterator without being marked as Iterable. However, it's evident that the actual Generator object created by your browser functions as both Iterable and Iterator (since it returns itself).

const g = gen();
g[Symbol.iterator]() == g; // true

Answer 2

In order to properly handle the series of values, it seems like using Iterable<T> instead of Iterator<T> is necessary. The interface Iterable contains the method [Symbol.iterator] that for-of loops call upon, returning an Iterator with a next() method to generate the values.

function* gen(): Iterable<number> {
  yield 1
  yield 2
  yield 3
}

Currently, TypeScript's handling of Generators appears to be incomplete. While there is a built-in type definition for Generator, it lacks generics (which means the produced values default to any). This would result in an error as it is identified only as an Iterator without being marked as Iterable. However, it's evident that the actual Generator object created by your browser functions as both Iterable and Iterator (since it returns itself).

const g = gen();
g[Symbol.iterator]() == g; // true

Answer 3

Answer №2

It is necessary to target a recent version of ES, use the --downlevelIteration flag, and include at least the "dom" and "esnext" libraries for this code to function correctly.

function* gen() {
    yield 1;
    yield 2;
    yield 3;
}

for (let val of gen()) {
    console.log(val);
}

Here is the transpiled demonstration:

"use strict";
var __generator = (this && this.__generator) || function (thisArg, body) {
    var _ = { label: 0, sent: function() { if (t[0] & 1) throw t[1]; return t[1]; }, trys: [], ops: [] }, f, y, t, g;
    return g = { next: verb(0), "throw": verb(1), "return": verb(2) }, typeof Symbol === "function" && (g[Symbol.iterator] = function() { return this; }), g;
    function verb(n) { return function (v) { return step([n, v]); }; }
    function step(op) {
        if (f) throw new TypeError("Generator is already executing.");
        while (_) try {
            if (f = 1, y && (t = op[0] & 2 ? y["return"] : op[0] ? y["throw"] || ((t = y["return"]) && t.call(y), 0) : y.next) && !(t = t.call(y, op[1])).done) return t;
            if (y = 0, t) op = [op[0] & 2, t.value];
            switch (op[0]) {
                case 0: case 1: t = op; break;
                case 4: _.label++; return { value: op[1], done: false };
                case 5: _.label++; y = op[1]; op = [0]; continue;
                case 7: op = _.ops.pop(); _.trys.pop(); continue;
                default:
                    if (!(t = _.trys, t = t.length > 0 && t[t.length - 1]) && (op[0] === 6 || op[0] === 2)) { _ = 0; continue; }
                    if (op[0] === 3 && (!t || (op[1] > t[0] && op[1] < t[3]))) { _.label = op[1]; break; }
                    if (op[0] === 6 && _.label < t[1]) { _.label = t[1]; t = op; break; }
                    if (t && _.label < t[2]) { _.label = t[2]; _.ops.push(op); break; }
                    if (t[2]) _.ops.pop();
                    _.trys.pop(); continue;
            }
            op = body.call(thisArg, _);
        } catch (e) { op = [6, e]; y = 0; } finally { f = t = 0; }
        if (op[0] & 5) throw op[1]; return { value: op[0] ? op[1] : void 0, done: true };
    }
};
var __values = (this && this.__values) || function (o) {
    var m = typeof Symbol === "function" && o[Symbol.iterator], i = 0;
    if (m) return m.call(o);
    return {
        next: function () {
            if (o && i >= o.length) o = void 0;
            return { value: o && o[i++], done: !o };
        }
    };
};
var e_1, _a;
function gen() {
    return __generator(this, function (_a) {
        switch (_a.label) {
            case 0: return [4 /*yield*/, 1];
            case 1:
                _a.sent();
                return [4 /*yield*/, 2];
            case 2:
                _a.sent();
                return [4 /*yield*/, 3];
            case 3:
                _a.sent();
                return [2 /*return*/];
        }
    });
}
try {
    for (var _b = __values(gen()), _c = _b.next(); !_c.done; _c = _b.next()) {
        var val = _c.value;
        console.log(val);
    }
}
catch (e_1_1) { e_1 = { error: e_1_1 }; }
finally {
    try {
        if (_c && !_c.done && (_a = _b["return"])) _a.call(_b);
    }
    finally { if (e_1) throw e_1.error; }
}

Answer 4

It is necessary to target a recent version of ES, use the --downlevelIteration flag, and include at least the "dom" and "esnext" libraries for this code to function correctly.

function* gen() {
    yield 1;
    yield 2;
    yield 3;
}

for (let val of gen()) {
    console.log(val);
}

Here is the transpiled demonstration:

"use strict";
var __generator = (this && this.__generator) || function (thisArg, body) {
    var _ = { label: 0, sent: function() { if (t[0] & 1) throw t[1]; return t[1]; }, trys: [], ops: [] }, f, y, t, g;
    return g = { next: verb(0), "throw": verb(1), "return": verb(2) }, typeof Symbol === "function" && (g[Symbol.iterator] = function() { return this; }), g;
    function verb(n) { return function (v) { return step([n, v]); }; }
    function step(op) {
        if (f) throw new TypeError("Generator is already executing.");
        while (_) try {
            if (f = 1, y && (t = op[0] & 2 ? y["return"] : op[0] ? y["throw"] || ((t = y["return"]) && t.call(y), 0) : y.next) && !(t = t.call(y, op[1])).done) return t;
            if (y = 0, t) op = [op[0] & 2, t.value];
            switch (op[0]) {
                case 0: case 1: t = op; break;
                case 4: _.label++; return { value: op[1], done: false };
                case 5: _.label++; y = op[1]; op = [0]; continue;
                case 7: op = _.ops.pop(); _.trys.pop(); continue;
                default:
                    if (!(t = _.trys, t = t.length > 0 && t[t.length - 1]) && (op[0] === 6 || op[0] === 2)) { _ = 0; continue; }
                    if (op[0] === 3 && (!t || (op[1] > t[0] && op[1] < t[3]))) { _.label = op[1]; break; }
                    if (op[0] === 6 && _.label < t[1]) { _.label = t[1]; t = op; break; }
                    if (t && _.label < t[2]) { _.label = t[2]; _.ops.push(op); break; }
                    if (t[2]) _.ops.pop();
                    _.trys.pop(); continue;
            }
            op = body.call(thisArg, _);
        } catch (e) { op = [6, e]; y = 0; } finally { f = t = 0; }
        if (op[0] & 5) throw op[1]; return { value: op[0] ? op[1] : void 0, done: true };
    }
};
var __values = (this && this.__values) || function (o) {
    var m = typeof Symbol === "function" && o[Symbol.iterator], i = 0;
    if (m) return m.call(o);
    return {
        next: function () {
            if (o && i >= o.length) o = void 0;
            return { value: o && o[i++], done: !o };
        }
    };
};
var e_1, _a;
function gen() {
    return __generator(this, function (_a) {
        switch (_a.label) {
            case 0: return [4 /*yield*/, 1];
            case 1:
                _a.sent();
                return [4 /*yield*/, 2];
            case 2:
                _a.sent();
                return [4 /*yield*/, 3];
            case 3:
                _a.sent();
                return [2 /*return*/];
        }
    });
}
try {
    for (var _b = __values(gen()), _c = _b.next(); !_c.done; _c = _b.next()) {
        var val = _c.value;
        console.log(val);
    }
}
catch (e_1_1) { e_1 = { error: e_1_1 }; }
finally {
    try {
        if (_c && !_c.done && (_a = _b["return"])) _a.call(_b);
    }
    finally { if (e_1) throw e_1.error; }
}

Answer 5

Answer №3

After some reflection, I have realized that the root cause of my issue was based on a mistaken assumption - believing that an Iterator could be directly used with for...of.

The underlying reason why the code snippet without type declarations functions in a web browser is because a return value from a generator can serve as both an Iterator and an Iterable. However, for the iteration process using for...of, only an Iterable object is acceptable according to the details provided on the MDN page.

When I explicitly specified the return type as an Iterator for the generator, it inadvertently led TypeScript to narrow its capabilities, causing it to overlook the inherent Iterable functionality.

A subsequent experiment utilizing solely an Iterator unsurprisingly failed in my browsing environment:

const it = {
  next: () => ({value: 1, done: false})
}

for (const val of it) {
  console.log(val)
}

Answer 6

After some reflection, I have realized that the root cause of my issue was based on a mistaken assumption - believing that an Iterator could be directly used with for...of.

The underlying reason why the code snippet without type declarations functions in a web browser is because a return value from a generator can serve as both an Iterator and an Iterable. However, for the iteration process using for...of, only an Iterable object is acceptable according to the details provided on the MDN page.

When I explicitly specified the return type as an Iterator for the generator, it inadvertently led TypeScript to narrow its capabilities, causing it to overlook the inherent Iterable functionality.

A subsequent experiment utilizing solely an Iterator unsurprisingly failed in my browsing environment:

const it = {
  next: () => ({value: 1, done: false})
}

for (const val of it) {
  console.log(val)
}

TypeScript does not accept an Iterator within a for...of loop

Answer №1

Answer №2

Answer №3

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