The distinction between TypeScript generics: Differences between a generic function type and the call signature of an object literal type (Type 'T' does not match type 'T')

I'm currently working on developing an RxJS operator that waits for the input operator to complete before switching to a second operator generated dynamically using switchMap. I have created two versions of the code, one that works perfectly and another that doesn't seem to function as expected. I am struggling to understand why there is a discrepancy between the two implementations.

The successful version is as follows:

import { Observable } from "rxjs"; // OperatorFunction,
import { defaultIfEmpty, last, switchMap } from "rxjs/operators";

// This definition of OperatorFunction is more or less equivalent to the
// definition in rxjs/src/internal/types.ts
interface OperatorFunction<T, S> {
    (input: Observable<T>): Observable<S>;
}

interface ObservableGenerator<T, S> {
    (value: T): Observable<S>;
}

export function switchMapComplete<T, S>(project: ObservableGenerator<T, S>): OperatorFunction<T, S> {
    function mapper(obs1: Observable<T>): Observable<S> {
        return obs1.pipe(
            defaultIfEmpty(null),
            last(),
            switchMap(project)
        );
    }

    return mapper;
}

The non-functional version, with changes only in the definitions of OperatorFunction and OperatorGenerator, looks like this:

import { Observable } from "rxjs";
import { defaultIfEmpty, last, switchMap } from "rxjs/operators";

type OperatorFunction2<T, S> = <T, S>(obs: Observable<T>) => Observable<S>;

type ObservableGenerator2<T, S> = <T, S>(value: T) => Observable<S>;

export function switchMapComplete2<T, S>(project: ObservableGenerator2<T, S>): OperatorFunction2<T, S> {
    function mapper(obs1: Observable<T>): Observable<S> {
        return obs1.pipe(
            defaultIfEmpty(null),
            last(),
            switchMap(project)
        );
    }

    return mapper;
}

The latter version results in a compiler error displaying the following exception:

error TS2322: Type 'Observable<{}>' is not assignable to type 'Observable<S>'.
  Type '{}' is not assignable to type 'S'.
util.ts(49,5): error TS2322: Type '(obs1: Observable<T>) => Observable<S>' is not assignable to type 'OperatorFunction2<T, S>'.
  Types of parameters 'obs1' and 'obs' are incompatible.
    Type 'Observable<T>' is not assignable to type 'Observable<T>'. Two different types with this name exist, but they are unrelated.
      Type 'T' is not assignable to type 'T'. Two different types with this name exist, but they are unrelated.

This result was unexpected, and despite referring to the TypeScript documentation which states that both versions should be equivalent, I encountered this issue.

If anyone can provide insights into why the equivalence between the two versions breaks down in this scenario, I would greatly appreciate it.

PS: If you require an RxJS operator similar to mine, here is an alternative solution which is simpler and leverages the existing types provided by RxJS:

import { Observable, ObservableInput, OperatorFunction, pipe } from "rxjs";
import { defaultIfEmpty, last, switchMap } from "rxjs/operators";

export function switchMapComplete<T, S>(project: (value: T) => ObservableInput<S>): OperatorFunction<T, S> {
    return pipe(
        defaultIfEmpty(null),
        last(),
        switchMap(project)
    );
}

Answer №1

To start off, it is recommended to modify the type

OperatorFunction2<T, S> = <T, S>(obs: Observable<T>) => Observable<S>
to simply
type OperatorFunction2 = <T, S>(obs: Observable<T>) => Observable<S>
since the outer T or S are not utilized in the definition of the type alias. The inner <T, S> overrides the outer names. Additionally, apply a similar adjustment to ObservableGenerator2.

It should be noted that

type F<T> = (x:T) => void
does not equate to type G = <T>(x:T)=>void. TypeScript does not permit entirely generic values. Type F represents a generic function that points to a specific function and requires a type parameter to operate (F is incorrect, F<string> is correct). On the other hand, Type G denotes a precise type relating to a generic function where G cannot accept a type parameter (G<string> is incorrect, G is valid). An instance of type F<string> is definite and only accepts string function inputs while a value of type G is flexible and can accommodate any input.

Although the types are not completely unrelated, they are distinct. Unfortunately, without having RxJS installed, errors may still exist in the following code. Nevertheless, the proposed solution is outlined below:

// concrete types referring to generic functions
type OperatorFunction2 = <T, S>(obs: Observable<T>) => Observable<S>;
type ObservableGenerator2 = <T, S>(value: T) => Observable<S>;

// a concrete function which takes a generic function and returns a generic function
export function switchMapComplete2(project: ObservableGenerator2): OperatorFunction2 {
  // a generic function
  function mapper<T, S>(obs1: Observable<T>): Observable<S> {
    return obs1.pipe(
      defaultIfEmpty(null),
      last(),
      switchMap(project)
    );
  }

  return mapper;
}

Hopefully, this guidance sets you on the right path. Further adjustments may be required within the implementation of mapper to rectify additional issues. Best of luck with your coding endeavors!

Answer №2

When working with generic function types in TypeScript, you have the flexibility to define them in different ways. For example:

type OperatorFunction2<T, S> = (obs: Observable<T>) => Observable<S>;

This syntax is commonly used for defining generic function types where the generic parameters are specified explicitly within the type definition.

However, another valid approach is:

type OperatorFunction2 = <T, S>(obs: Observable<T>) => Observable<S>;

In this style, the generic parameters are declared inline within the function signature. While this method can be useful in certain situations, it's important to note that it results in two independent sets of generic parameters with the same names T and S.

To combine both approaches, you can do:

type OperatorFunction2<T, S> = <T, S>(obs: Observable<T>) => Observable<S>;

By following these guidelines, you can effectively define and use generic function types in TypeScript to suit your specific requirements.

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