Alternative for __dirname in node when using the --experimental-modules flag

Question

I use the flag --experimental-modules when running my node application in order to use ES6 modules.

However when I use this flag the metavariable

Answer 1

As Geoff pointed out the following code returns not the module's path but working directory.

import path from 'path';
const __dirname = path.resolve();

works with --experimental-modules

Answer 2

There have been proposals about exposing these variables through import.meta, but for now, you need a hacky workaround that I found here:

// expose.js
module.exports = {__dirname};

// use.mjs
import expose from './expose.js';
const {__dirname} = expose;

Answer 3

process.cwd()

From documentation:

The process.cwd() method returns the current working directory of the Node.js process.