TypeScript - 分布式条件类型

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;} type Result1 = { A: any;} | { B: any;} .

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

Playground Link

Answer 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