How to alias complex type constructor in typescript? - oop

Question

I am using a custom library, and for some types I have to write:
import * as pipeline from 'custom-pipeline';
import {MapTransform} from 'custom-pipeline';
export const createTransform = (appConfig: IAppConfig):
MapTransform<ISomeData, IFoo> |
MapTransform<ISomeData, IBar> => {
switch(appConfig.createType) {
case 'foo':
return new pipeline.MapTransform<ISomeData, IFoo>((data: ISomeData) => {...});
case 'bar':
return new pipeline.MapTransform<ISomeData, IBar>((data: ISomeData) => {...});
}
}
Especially the verbose constructor is irking me. I can I alias the types alright:
type FooTransform = MapTransform<ISomeData, IFoo>;
type BarTransform = MapTransform<ISomeData, IBar>;
Yet I cannot do:
new FooTransform((data: ISomeData) => {...});
new BarTransform((data: ISomeData) => {...});
throwing an error like:
error TS2304: Cannot find name 'FooTransform'.
I assume it's because I only have a type and not a class? Yet how can I alias the constructor in a way that I can do new FooTransform as above?
The definition of MapTransform looks like:
export declare class MapTransform<A, B> extends BaseTransform<A, B> {
constructor(private mapFunc: (val: A) => B);
}
I can reduce the constructor to this:
fooMapFunction = (data: ISomeData): IFoo => {...};
new MapTransform<ISomeData, IFoo>(mapFunction);
Albeit it's not en par with a new FooTransform(fooMapFunction).

Answer 1

Your assumption is correct, type declarations are compile-time only, thus you can't perform any operations on them (including instantiation with new).
Solution 1: Type Assertion
Let's assume the superclass looks like this:
class MapTransform<T1> {
constructor(public readonly data: T1) { /* ... */ }
}
To create a specialised type alias, you can do this:
type TransformA = MapTransform<number>;
From the perspective of the superclass MapTransform, there's no difference between MapTransform<A>, MapTransform<B> and so on (that's the point of generics), so we can safely assign the constructor function of class MapTransform to a constant TransformA. Calling new TransformA() is, at runtime, identical to calling new MapTransform<number>():
const TransformA = <{ new (data: number): TransformA; }>MapTransform;
Notice the type assertion? This tells the TypeScript compiler to treat the assigned value MapTransform as an object which constructs an object of type TransformA when instantiating it with new. We can now write:
const a = new TransformA(123);
a.data.toExponential(); // works
BTW, what the TypeScript compiler actually sees is this ...
const TransformA = <{ new (data: number): MapTransform<number>; }>MapTransform;
... because type TransformA ≡ MapTransform<number>.
Be aware that all of these will evaluate true:
new TransformA(123) instanceof TransformA
new TransformA(123) instanceof MapTransform
new MapTransform<number>(123) instanceof TransformA
new MapTransform<number>(123) instanceof MapTransform
Here's the example in the TypeScript Playground.
Solution 2: Subclassing
Again, let's assume the superclass looks like this:
class MapTransform<T1> {
constructor(public readonly data: T1) { /* ... */ }
}
With subclassing, the solution is simple: Extend the superclass and pass the desired type parameters along in the extends clause.
class TransformA extends MapTransform<number> { }
Et voilà, you now have a constructor that works at runtime as well as a type that works at compile-time.
Unlike the first solution, the following 2 expressions will evaluate true ...
new TransformA(123) instanceof TransformA
new TransformA(123) instanceof MapTransform
... while these will evaluate false:
new MapTransform<number>(123) instanceof TransformA
new MapTransform<number>(123) instanceof MapTransform
Solution 2.1: Anonymous Subclassing
If you only need the constructor alias but not the type alias, this might come in handy:
const TransformA = class extends MapTransform<number> { };
This is called a class expression and can be used like every other expression, for example:
class App {
private static TransformA = class extends MapTransform<number> { };
public doStuff() {
return new App.TransformA(123);
}
}
More on Types
If you're interested, here's a few more links on the topic:
Named Types in TypeScript Spec
Type Assertions
TypeScript Handbook: Interfaces, specifically section "Difference between the static and instance sides of classes"
Class Expressions
EDIT
You wrote you've had issues applying these solutions to a class that expects a function as a parameter, so here's another example in the TypeScript playground.