How to export only one class (which uses another class) in TypeScript (CommonJS)?

Question

I implemented a "Least Recently Used Cache" (LRU Cache) using a doubly linked list. The linked list consists of nodes (Node class) which are used within a LRUCache class.

To export my module I am using the following pattern:

export class Node { ... }
export class LRUCache { ... }

I am exporting my module as a CommonJS module. With the export class pattern above, I am able to use my module the following way:

var module = require('lru-cache');
var cache = new module.LRUCache();

That's okay, but I don't like the additional "module." part in front of LRUCache. I would like to use my CommonJS exported module like this:

var LRUCache = require('lru-cache');
var cache = new LRUCache();

So I changed my code into this:

class Node { ... }
class LRUCache { ... }
export = LRUCache;

But now TypeScript complains with:

error TS4055: Return type of public method from exported class has or is using private name 'Node'.

It's because LRUCache uses the Node class and Node is not exported when defining export = LRUCache;.

How can I solve this situation? I am using TypeScript 2.1.

Screenshot:

Answer 1

With typescript you can simply do:

import { LRUCache } from "lru-cache";
let cache = new LRUCache();

And it will compile into:

var lru_cache_1 = require("./lru-cache");
var cache = new lru_cache_1.LRUCache();

If you prefer to use require, you can:

class Node { ... }
class LRUCache { ... }

export = LRUCache;

And then:

import LRUCache = require('lru-cache');
var cache = new LRUCache();

You can find much more info on this in the following two doc pages:

• Module Resolution
• Modules