Advanced TypeScript Exercises - Question 4

#typescript #javascript

For given function type F, and any type A (any in this context means we don't restrict the type, and I don't have in mind any type 😉) create a generic type which will take F as first argument, A as second and will produce function type G which will be the same as F but with appended argument A as a first one.

// lets say we have some function type
type SomeF = (a: number, b: string) => number
// and we have our utility type
type AppendArgument<F, A> = ... here your code 💪

type FinalF = AppendArgument<SomeF, boolean> 
// FinalF should be (x: boolean, a: number, b: string) => number

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Top comments (7)

Rahul Kashyap • Feb 20 '20

// lets say we have some function type
type SomeF = (a: number, b: string) => number

// and we have our utility type
type AppendArgument<F extends (...args: any[]) => any, A> = (x: A, ...args: Parameters<F>) => ReturnType<F>

type FinalF = AppendArgument<SomeF, boolean> 
// FinalF should be (x: boolean, a: number, b: string) => number

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ChenLee • Apr 8 '22

You should keep the arguments' order, so x must after a and b.

Jasper De Sutter • Feb 25 '20

// lets say we have some function type
type SomeF = (a: number, b: string) => number
// and we have our utility type
type AppendArgument<F, A> = F extends (...args: infer Args) => infer R ? (x: A, ...args: Args) => R : never

type FinalF = AppendArgument<SomeF, boolean> 
// FinalF should be (x: boolean, a: number, b: string) => number

John Ralph Umandal • Mar 9 '20 • Edited

// lets say we have some function type
type SomeF = (a: number, b: string) => number;
// and we have our utility type
type AppendArgument<F, A> =
    F extends (...[a, b, ...oth]: [infer ArgA, infer ArgB, infer ArgOth]) => infer FReturn
    ? (x: A, ...[a, b, ...oth]: [ArgA, ArgB, ArgOth]) => FReturn
    : unknown;

type FinalF = AppendArgument<SomeF, boolean> 
// FinalF should be (x: boolean, a: number, b: string) => number

I wasn't able to figure out myself how to dynamically take the pair arg:type out from a generic type function :(

UPDATE: palmface

// lets say we have some function type
type SomeF = (a: number, b: string) => number;
// and we have our utility type
type AppendArgument<F, A> =
    F extends (...args: infer Args) => infer FReturn
    ? (x: A, ...args: Args) => FReturn
    : unknown;

type FinalF = AppendArgument<SomeF, boolean> 
// FinalF should be (x: boolean, a: number, b: string) => number

David Johnston • Apr 7 '20 • Edited

Just a note - the term 'append' means to 'add to the end of' (that's why the appendix is at the end of the book). The term you should be using here is `prepend'.