Typescript Functional Type Inference

Question

I've written this simple compose function which works just fine. However in order to assure type safety I had to resort to using generics to provide the compiler with type hints even though it's easily possible to infer the signature of "upperCaseAndLog".

const compose = <T, R>(...fns: Array<(a: any) => any>) => (a: T): R =>
  fns.reduce((b, f) => f(b), a);

const greet = (s: string) => "Hello " + s;
const toUpperCase = (s: string) => s.toUpperCase();
const log = console.log;

const upperCaseAndLog = compose<string, void>(
  greet,
  toUpperCase,
  log
);

upperCaseAndLog("bill");

Am I missing something, is there a more elegant way of accomplishing the same goal? I assume that languages like F# or Haskell would be able to infer the types without any additional information.

The compiler cannot infer your types because they don't appear in the parameters. How do you expect it to be able to tell? — SLaks

Titian Cernicova-Dragomir Titian Cernicova-Dragomir · Accepted Answer · 2018-11-07T07:01:13

Typescript can't infer such linked types (linked in the sense that the argument o the function depends on the result of the previous function).

You can't even define the signature of compose in a general enough way that it works for an number of function. What we can do, is define overloads that accept up to a given number of functions:

type Fn<A, R> = (a: A) => R // just to be a bit shorter in the compose signature, you can use teh function signature directly  
function compose<T, P1, P2, R>(fn1: Fn<T, P1>, fn2: Fn<P1, P2>, f3: Fn<P2, R>) : Fn<T, R>
function compose<T, P1, R>(fn1: Fn<T, P1>, f2: Fn<P1, R>) : Fn<T, R>
function compose(...fns: Array<(a: any) => any>) {
    return function (a: any) {
        return fns.reduce((b, f) => f(b), a);
    }
}

const greet = (s: string) => "Hello " + s;
const toUpperCase = (s: string) => s.toUpperCase();
const log = console.log;

const upperCaseAndLog = compose(
    greet,
    toUpperCase,
    log
);

upperCaseAndLog("bill");//(a: string) => void

Typescript Functional Type Inference

1 Answers