CODEWITHSUNDEEP
typescripttypesoptional-parametersconditional-types
eczn
eczn

Reputation: 1801

why TypeScript condition type can't work correctly for optional parameter?

i want to know wheather function's first args is optional, so i write this:

type F1 = (payload?: number) => null;
type Res1 = F1 extends (a?: any) => any ? 1 : 2; // got 1
type Res2 = F1 extends (a: any) => any ? 1 : 2; // got 1, but 2 expected

type F2 = (payload: number) => null;
type Res3 = F2 extends (a?: any) => any ? 1 : 2; // got 1, but 2 expected
type Res4 = F2 extends (a: any) => any ? 1 : 2; // got 1

ALL OF THEM are both equl to 1, why ? and how do i do ?

Upvotes: 0

Views: 131

Answers (1)

mbdavis
mbdavis

Reputation: 4010

You can read 'extends' as 'is assignable to'.

type F1 = (payload?: number) => null;
type Res2 = F1 extends (a: any) => any ? 1 : 2; // got 1, but 2 expected

Is (payload?: number) => null assignable to (a: any) => any ? Let's check:

// Works
const res2: (a: any) => any = (payload?: number) => null;

(a: any) => any accepts any value for a, including undefined or a number as specified by payload, so we can see that (payload?: number) => null is assignable and therefore F1 extends (a: any) => any should evaluate to true and return 1.

Instead, you can make the arguments of your function to test optional using Partial, and check if the resulting arguments are assignable to the original arguments:


type ArgumentType<F extends (...args: any) => any> = F extends (...args: infer A) => any ? A : never;

type FirstArgumentOptional<F extends (...args: any) => any> =
    Partial<ArgumentType<F>> extends ArgumentType<F>
        ? true 
        : false


type T1 = FirstArgumentOptional<(a: any) => any>; // false
type T2 = FirstArgumentOptional<(a?: any) => any>; // true
type T3 = FirstArgumentOptional<() => any>; // true

Upvotes: 1

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