Suggestion: consolidate generics in union if they match

TypeScript Version: 3.5.2

Search Terms: rxjs startWith generics union consolidate same

Code

declare const create: <T>(
    t: T,
) => {
    startWith: <D>(v1: D) => T | D;
};

// Succeeds
// $ExpectType { foo: number } | { bar: number }
const r1 = create({ foo: 1 }).startWith({ bar: 1 });

// Fails
// $ExpectType { foo: number }
// Actual: { foo: number } | { foo: number }
const r2 = create({ foo: 1 }).startWith({ foo: 1 });

Although the expected/actual types are compatible, it's slightly irritating to deal with a union type T | T instead of a consolidated type T, as it makes type inspection in VS Code much more difficult (especially when the types are large or deeply nested).

Real world use case: https://github.com/ReactiveX/rxjs/issues/4866

首先使用 TypeScript 3.5.2 重现提供的 create/startWith 示例，并检查 r1 和 r2 的推断类型。调查负责处理相同泛型结果的联合类型的类型检查路径；当匹配的联合类型表示为单个类型且不改变不同联合类型的情况时，即视为完成。