请考虑以下事项:
type U = {a: string, b: string};
type T = U & ({c: string, d?: never}|{c?: never, d: string});
type E = Exclude<T, {a: string}>;
type O = Omit<T, "a">;
const t1: T = {a: "a", b: "b", c: "c", d: "d"}; // error - expected
const t2: T = {a: "a", b: "b", c: "c"}; // ok - expected
const t3: T = {a: "a", b: "b", d: "d"}; // ok - expected
const e1: E = {b: "foo", c: "c", d: "d"}; // error - expected
const e2: E = {b: "foo", c: "c"}; // error: E is never
const e3: E = {b: "foo", d: "d"}; // error: E is never
const o1: O = {b: "foo", c: "c", d: "d"}; // ok - unexpected!
const o2: O = {b: "foo", c: "c"}; // ok
const o3: O = {b: "foo", d: "d"}; // ok
类型T被定义为包含属性a和b的对象and或c或d，但not两者都包含.
我正在try 构造一个与T except相同的类型，它没有a属性.我try 了Exclude和Omit，从上面可以明显看出，这两种方法都不能满足我的需求.
有没有一种方法可以在不事先知道类型T的 struct 的情况下，从类型T的黑盒中删除一个属性，并保持它的其他所有内容不变？
还打字here个操场.
Edit:修复了原始类型；添加了更多测试用法
Edit #2:这道题与Typescript: Omit a property from all interfaces in a union, but keep the union structure题不同，因为:
- 所讨论的类型不是直线并集，而是一侧包含并集的交集
- 它还解决了从任意"黑盒"类型中删除的属性而不更改其任何其他语义的问题.