Optional parameters in typed languages can either have zero or all parameters provided

Question

Optional parameters in typed languages can either have zero or all parameters provided

I am delving into the world of typed functional programming and have embarked on implementing partial application with a focus on type safety.

Issue at hand: I'm aiming to create a function that can take a function along with zero or all of its parameters as arguments.

Here is the initial interface:

interface Functor {
  (...args: any[]) => any
}

This led me to develop the following function:

const partial = <T extends Functor>(fx: T, ...apply: Parameters<T>): Functor =>
  (...args: any[]) => fx(...apply, ...args);

The dilemma lies in ...args: Parameter<T>, which prompts typescript to expect all parameters whereas I want to allow zero up to all.

Is there a workaround for this particular issue?

typescript functional-programming partial-application

Answer 1

Answer №1

To create a utility called PartialParameters that works similar to the built-in Parameters:

type PartialParameters<T> = T extends (...args: infer P) => any ? Partial<P> : never;

const partial = <T extends Functor>(fx: T, ...apply: PartialParameters<T>): Functor =>
(...args: any[]) => fx(...apply, ...args);

declare function foo(a: string, b: number): boolean;

partial(foo)

partial(foo, '1')

partial(foo, '1', 1)

Interactive Example

Answer 2

To create a utility called PartialParameters that works similar to the built-in Parameters:

type PartialParameters<T> = T extends (...args: infer P) => any ? Partial<P> : never;

const partial = <T extends Functor>(fx: T, ...apply: PartialParameters<T>): Functor =>
(...args: any[]) => fx(...apply, ...args);

declare function foo(a: string, b: number): boolean;

partial(foo)

partial(foo, '1')

partial(foo, '1', 1)

Interactive Example

Optional parameters in typed languages can either have zero or all parameters provided

Answer №1

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