TypeScript - Distributive conditional types

Question

Talk is cheap,show the code! By the way, the version of ts I am testing is 4.6.4

type ITypeA = ((args: { A: any }) => any) | ((args: { B: any }) => any);

type Test<T> = T extends (args: infer A) => any ? A : never;

// type Result1 = {
//     A: any;
// } | {
//     B: any;
// }
type Result1 = Test<ITypeA>;

// type Result2 = {
//     A: any;
// } & {
//     B: any;
// }
type Result2 = ITypeA extends (args: infer A) => any ? A : never;

Result1 may use 'distributive conditional types' rule in ts, so type Result1 = { A: any;} | { B: any;}.

My question is why does Result2 not apply this rule? Is there any difference between them?

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captain-yossarian from Ukraine · Accepted Answer · 2022-05-26 10:29:06Z

1

Please see docs

When conditional types act on a generic type, they become distributive when given a union type. For example, take the following:

There is a requirement, you need act on generic.

Result2 acts on known types, there is no generic, whereas Result1 uses Test which in turn uses generic T

answered May 26, 2022 at 10:29

captain-yossarian from Ukraine

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1 Comment

meisu Over a year ago

Thank you!! I understand! It must act on generic!

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