Typescript：将function的参数类型定义为function和Z56B97998B338B9F37776

Question

I have created a compose function like this

const composeTyped = <T, U, R>(f: (x: T) => U, g: (y: U) => R) => (x: T) => g(f(x));

It seems to me that both f and g are functions of type fGeneric which is defined as

type fGeneric = <T, R>(arg: T) => R;

My problem is that I do not understand if and how I could use the type fGeneric to specify the type of f and g in composedType . To be more clear, if I do like this

const composeTyped_ = <T, U, R>(f: fGeneric, g: fGeneric) => (x: T) => g(f(x));

the function composeTyped_ is assigned the type (x: T) => unknown . What I would like to obtain though is the type (x: T) => R .

Answer 1

You need to define fGeneric so that it accepts generic type arguments:

type fGeneric<T, R> = (arg: T) => R;

Then you can defined composeTyped_ like so:

const composeTyped_ = <T, U, R>(
  f: fGeneric<T, U>,
  g: fGeneric<U, R>
) => (x: T) => g(f(x));

This should now work nicely:

declare const f: (str: string) => number

declare const g: (num: number) => null

composeTyped_(f, g)

// Argument of type '(num: number) => null' is not assignable to parameter of type 
//   'fGeneric<number, string>'.
//  Type 'null' is not assignable to type 'string'.
composeTyped_(g, f)