Skip to content

Allow inferring rest element types in conditional types involving tuples #25719

Open

Description

@AlCalzone

Search Terms

rest element infer tuple

Suggestion

Currently, inferring single elements of tuple types is possible using the infer keyword:

type FirstArg<T extends any[]> = 
	T extends [infer R, ...any[]] ? R : 
	T extends [] ? undefined : 
	never;
type T1 = FirstArg<[number, 2, 3]>; // number
type T2 = FirstArg<[1, 2, 3]>; // 1
type T3 = FirstArg<["", 2]>; // ""
type T4 = FirstArg<[]>; // undefined

However it is not possible to infer the type of the remaining arguments in one go, except by resorting to functions:

type RestArgs<T extends any[]> = 
	T extends [any, infer R[]] ? R : // this does not work - no way to specify that R should be an array!
	T extends [any] ? []] : 
	never;

// this does
type RestArgs<T extends any[]> =
	((...args: T) => void) extends ((first: any, ...rest: infer S1) => void) ? S1
	: T extends [infer S2] ? []
	: T extends [] ? []
	: never;
type T1 = RestArgs<[1,2,3]>; // [2, 3]
type T2 = RestArgs<[1,2]>; // [2]
type T3 = RestArgs<[1]>; // []
type T4 = RestArgs<[]>; // []

I would like to see the possibility to infer rest types in tuples, e.g. like this (square brackets):

type RestArgs<T extends any[]> = T extends [any, infer R[]] ? R : never;

or like this (3 dots)

type RestArgs<T extends any[]> = T extends [any, infer ...R] ? R : never;

Checklist

My suggestion meets these guidelines:

  • This wouldn't be a breaking change in existing TypeScript / JavaScript code
  • This wouldn't change the runtime behavior of existing JavaScript code
  • This could be implemented without emitting different JS based on the types of the expressions
  • This isn't a runtime feature (e.g. new expression-level syntax)

Activity

  1. jscheiny commented on Jul 25, 2018

    @jscheiny

    I just ran into this issue myself and wanted to throw in my two cents about the syntax for how this would behave. Considering how this inference is written for function types and how rest elements of tuples are constructed:

    type Tail<T> = T extends (head: any, ...tail: infer U) ? U : never;
    
    type Rest = string[];
    type Tuple = [any, ...Rest];

    I would suggest the following syntax:

    type Tail<T> = T extends [any, ...infer U] ? U : never;
  2. aleclarson commented on Oct 27, 2018

    @aleclarson

    The current workaround:

    type Tail<T extends any[]> = ((...args: T) => any) extends ((
        _: infer First,
        ...rest: infer Rest
    ) => any)
        ? T extends any[] ? Rest : ReadonlyArray<Rest[number]>
        : []

    Note: Does not work in TypeScript 2.9.x and under.

  3. darcyparker commented on Jun 11, 2019

    @darcyparker

    AlCalzone - Thanks for the examples FirstArg and RestArgs. They inspired these additional examples:

    export type Length<T extends any[]> = T extends (infer U)[] & { length: infer L } ? L : never;
    export type LengthMinusOne<T extends any[]> = Length<Tail<T>>;
    export type LengthPlusOne<T extends any[]> = Length<Cons<any, T>>;
    export type Last<T extends any[]> = T[LengthMinusOne<T>];
    // Example:
    // export type TestLastType = Last<[boolean, string, number]>; //number

    Based on the above, it would be great to have something like:

    type Last<T extends any[]> = T extends [...any[], infer L] ? L : never;
  4. smcatala commented on Aug 14, 2019

    @smcatala
    type Tail<A extends any[]> = 
      ((...args: A) => any) extends ((h: any, ...t: infer T) => any) ? T : never
  5. chocolateboy commented on Nov 14, 2020

    @chocolateboy

    This works in TypeScript v4 (via variadic tuple types):

    type Tail<T extends any[]> = T extends [any, ...infer U] ? U : never;
  6. ethanresnick commented on Sep 19, 2023

    @ethanresnick
    Contributor

    This seems like it can probably be closed, given the snippet shown above?

  7. futtetennista commented on Jan 29, 2024

    @futtetennista

    Not sure that works but I might be missing something here. For example

    type Tail<T extends any[]> = T extends [any, ...infer U] ? U : never;
    const xs = ["1", "2", "3"];
    type T = Tail<typeof xs>; // never

    Playground link

  8. ethanresnick commented on Jan 29, 2024

    @ethanresnick
    Contributor

    Stefano (@futtetennista) It fails in your example because typeof xs is not a tuple type (it's string[]). Here is an updated version that works:

    const xs = ["1", "2", "3"] as const
    type Xs = typeof xs
    type Tail<T extends readonly any[]> = T extends readonly [any, ...infer U] ? U : never;
    type TT = Tail<Xs>
Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment

Metadata

Metadata

Assignees

No one assigned

    Labels

    In DiscussionNot yet reached consensusSuggestionAn idea for TypeScript

    Type

    No type

    Projects

    No projects

      Milestone

      No milestone

      Relationships

      None yet

      Development

      No branches or pull requests

      Issue actions