ts-no-new

typescript-constructors-without-new

Various attempts to create TypeScript constructors that can be invoked with and without the new keyword like several built-ins including Array et al.

Feedback

I am not 100% happy with any of those solutions. Any feedback with better ideas is more than welcome.

I would especially like to:

1. eliminate all any types (without making the proxy wrapper specific to one particular type - which is tricky, if possible at all)
2. eliminate proxies (currently I don't think it's possible but hopefully I'm wrong)

I will appreciate any help with that.

Background

This is my old code that I was playing some time ago in 2018. I'm putting it online in case it is useful to someone.

All examples is in a separate directory with its own tsconfig.json file to avoid errors of the same symbols defined in multiple files.

The tsconfig.json is as strict as possible, with all options turned on that I usually recommended. It made it harder to write the code but it is at the same time more useful in different compiler configurations.

The idea is to match the built-in costructors like Array as closely as possible. There are several important things to keep in mind:

1. It should be all correct TypeScript with strict type checking
2. It should use class keyword instead of defining the constructors as old style functions
3. The compiler should understand the types and the VSCode editor should correctly detect errors like a typo in a method name exactly the same as normal constructors with new keyword in TypeScript - no type safety compromises
4. Both a = new A() and a = A() should work the same
5. In both cases a instanceof A should be true
6. In both cases a.constructor.name should be 'A'
7. In both cases console.log(a) should print A {...}
8. Another constructor B should be able to extend (inherit from) A using class B extends A { ... } syntax
9. Both b = new B() and b = B() should both work
10. In both cases b instanceof B and b instanceof A and should be true
11. In both cases a.constructor.name ahould be 'B'
12. In both cases console.log(a) should print B {...}

Optional requirement (is it important? or can it be harmful?):

1. In both cases for the base class (a = new A() and a = A()) a.constructor == A should be true
2. In both cases for the child class (b = new B() and b = B()) b.constructor == B should be true

Can it even be done without proxying the instances or changing the costructor to redefine the constructor property of the object? (which can be made non enumerable but would still return true for hasOwnProperty) It may not be worth the hassle but I may be wrong here.

If you have some other important feature that is not listed here, please post an issue.

Usage

The exampels are in separate directories. The rsp-2-* ones are most useful. All of the rsp-0-* and rsp-1-* as well as the bak* directories are mostly for backup archives but feel free to explore my early attempts (but not everything can work).

The only dependency is the TypeScript compiler.

If you don't have the TypeScript compiler installed globally then install it locally with:

npm i

and use it as npx tsc as shown below. If you have the TypeScript compiler installed globally then you can use tsc instead of npx tsc in the examples below.

To run the rsp-2-4 example:

cd rsp-2-4
npx tsc
node build/*.js

or:

./run rsp-2-4

It will print some output that is relevant to the required features of the given solution. If you need more info just add it to the source code.

Issues

For any bug reports or feature requests please post an issue on GitHub.

Author

Rafał Pocztarski

License

MIT License (Expat). See LICENSE.md for details.