TypeScript generics: how to define type T which is structurally the same as other type S

Question

I have a notion of a Step , which requires value of type A as an input and gives out a value of type B .

class Step<A, B> {
  constructor(private readonly f: (a: A) => B) { }

  public run(a: A): B {
    return this.f(a);
  }
}

Now I would like to compose two steps, so I end up with something like this inside class Step :

  public andThen<C, D>(nextStep: Step<C, D>): Step<A, D> {
    return new Step<A, D>((state: A) => {
      const b: B = this.f(state);
      return nextStep.run(b);  // <---- compile error, B and C have no relation defined
    });
  }

What I would like to achieve is to somehow tell the type system that we can pass type B to a function which expects type C (structural typing should check that all fields in C are present in B ), so that the line return nextStep.run(b) works fine.

Example:

const stepA: Step<{}, {a: number, b: string}> = new Step((input: {}) => ({ a: 5, b: "five" }));
const stepB: Step<{a: number}, {c: number}> = new Step((input: {a: number}) => ({c: input.a + 5}));

const steps = stepA.andThen(stepB)

As you can see stepB requires as an input {a: number} , so it can be fed an output from stepA which is {a: number, b: string} . But I cannot figure out how to define the relation in andThen . Any thoughts how this can be achieved?

Answer 1

type SubType<T, WiderT> = T extends WiderT ? WiderT : never;

class Step<A, B> {
  constructor(private readonly f: (a: A) => B) { }

  public run(a: A): B {
    return this.f(a);
  }

  public andThen<C, D>(nextStep: Step<SubType<B, C> | B, D>): Step<A, D> {
    return new Step<A, D>((state: A) => {
      const b = this.f(state);
      return nextStep.run(b);
    });
  }
}

const stepAB: Step<{}, {a: number, b: string}> = 
new Step((input) => ({ a: 5, b: "five" }));

const add5 = ({ a }: { a: number }) => ({ c: a + 5 });
const stepCD: Step<{ a: number }, { c: number }> = new Step(add5);


const stepAD = stepAB.andThen(stepCD);

The core point is SubType type:

type SubType<T, WiderT> = T extends WiderT ? WiderT : never;

It says that if the type extends given type then its ok, but its not then never. Pay attention how it is used:

SubType<B, C> | B

So we are saying that if type B extends C , so B is more specific type then C then we allow on that, if not we don't by never . Additionally we allow on B itself (without union there was still compilation error)