function extend<T, U> (first: T, second: U): T & U {
  let result = {} as T & U
  for (let id in first) {
    result[id] = first[id] as any
  }
  for (let id in second) {
    if (!result.hasOwnProperty(id)) {
      result[id] = second[id] as any
    }
  }
  return result
}

1. first: T, second: U中的T,U是代表first,second本身还是内部所有成员？
2. result[id] = first[id] as any，这段代码result 接受的是T & U类型,但为什么first[id]类型断言为any, 却可以通过，这是为什么？
3. 为什么这种写法会报错，first[id]改为first[id] as any就可以通过？

function extend<T, U> (first: T, second: U): T & U {
  let result = {} as T & U
  for (let id in first) {
    result[id] = first[id]
  }
  for (let id in second) {
    if (!result.hasOwnProperty(id)) {
      result[id] = second[id]
    }
  }
  return result
}