What is the significance of using a polyfill with TypeScript?

If I want to utilize Object.values(), I need to designate

"lib":["es2017"]
. However, my
"target":"es6"
.

My interpretation is that I am using methods from es2017 but the result is es6 code.

Considering this, why would a polyfill be necessary and how can I identify a trustworthy one?

Answer №1

TypeScript differentiates between API invocations and syntax structure.

The TypeScript compiler breaks down the syntax elements (identified by special symbols like =>, ?, `, # and keywords such as class or static) but does not handle the API itself. To emulate API calls, like Array.prototype.flat (introduced in ES2019), you would require an additional compiler like Babel.

Illustration

The below compiler configuration will convert the nullish coalescing operator (??) and the class declaration but it won't directly alter the API call to Array.prototype.flat (as it is not supported in ES5):

tsconfig.json

{
  "compilerOptions": {
    "lib": ["ES2019"],
    "target": "ES5"
  }
}

main.ts

export class MyConverter {
  static flatten(numbers?: (number | number[])[]) {
    return (numbers ?? []).flat();
  }
}

main.js

"use strict";
Object.defineProperty(exports, "__esModule", { value: true });
exports.MyConverter = void 0;
var MyConverter = /** @class */ (function () {
    function MyConverter() {
    }
    MyConverter.flatten = function (numbers) {
        return (numbers !== null && numbers !== undefined ? numbers : []).flat();
    };
    return MyConverter;
}());
exports.MyConverter = MyConverter;

Answer №2

Typescript is designed to have minimal runtime impact, with only a few utility functions causing any additional runtime behavior.

Since Typescript cannot predict the environment in which your code will run or the level of support for certain features, specifying a lib or target option informs Typescript that there will be runtime support for those features without concern for how it will be implemented.

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