How to define a generic anonymous function type in TypeScript?

Question

Is it possible to have Generic Anonymous function type?

I was reading this article and found this piece of code.

import { Eq } from 'fp-ts/Eq'

export const not = <A>(E: Eq<A>): Eq<A> => ({
  equals: (first, second) => !E.equals(first, second)
})

Is not function here even a valid typescript syntax?

Yup .. it's valid./. not sure what the question is: typescriptlang.org/play?#code/… — Titian Cernicova-Dragomir
– Titian Cernicova-Dragomir, Commented Aug 4, 2022 at 8:45
@TitianCernicova-Dragomir Thank you. In your example in the playground you used comma in your generic type parameter <A,>. Without it it's seems to be a syntax error. What is that comma for? — Shnd
– Shnd, Commented Aug 5, 2022 at 5:03
In TSX files <T> is intrepreted as a JSX tag instead of a generic parmater list. The , forces it to be a generic parameter list. The playground works in TSX mode — Titian Cernicova-Dragomir
– Titian Cernicova-Dragomir, Commented Aug 5, 2022 at 16:50

Matthieu Riegler · Accepted Answer · 2022-08-05 07:02:17Z

1

This code is perfectly fine ,

It's (almost) the equivalent of this generic function but with an arrow function definition :

function not2<A>(E: Eq<A>): Eq<A> {
  return {
    equals: function (first, second) {
       return !E.equals(first, second);
    }
  };
}

answered Aug 4, 2022 at 8:17

Matthieu Riegler

59.9k27 gold badges158 silver badges208 bronze badges

Sign up to request clarification or add additional context in comments.

1 Answer 1