欺骗类型记录编译器将错误接口分配给变量。

Question

我试图通过滥用类型记录来强制执行自定义类型规则。我让它与类一起工作，但是我在接口方面遇到了困难。下面是工作的类版本：

export class HeaderElement extends WebElement { 
  // 'dummy' type: won't create any instances of it at runtime.
  // type-filler member that doesn't exist in other WebElement.
  _HEADER_ELEMENT_TYPE_FILLER_6897689FSAFDSF: string; 
  //  -> dummy member: not even read/written/set in code.
}
const export function getHeaderElement(): HeaderElement {
  // findElement: returns WebElement type
  const header = browser.findElement( ... );
  return header as HeaderElement; 
  // **tricking typescript (actual runtime type would be 'WebElement')**
}
export function clickWowPageMenu(
  headerElem: HeaderElement
  // (headerElem's runtime-type is WebElement. 
  // It's 'HeaderElem' only in compilation phase.)
) { ... }
const headerElem = getHeaderElement();
const saveButtonElem = getSaveButtonElement();
clickWowPageMenu( headerElem ); // ok
clickWowPageMenu( saveButtonElem ); // type-error.

但是对于接口版本，以下代码会导致类型记录类型错误：

interface IContext {
   __EXTENDS_IContextElem123124: string,
}
interface IChatContext extends IContext {
   name: string,
}
const test: IChatContext = { name: 'namehere' } as IChatContext;
// want: no type error here. (but 'missing __EXTENDS_IContextElem123124')
const test2: IChatContext = {} as IChatContext;
// want: type error here. (missing property 'name')

我如何“滥用”类型记录，以我想要的方式输入检查界面？或者还有别的办法(如。官方的)？

谢谢!

const test: IContextElem ... --> IContext

1. 添加缺失的详细信息(想要/不想要)

Answer 1

TypeError是正确的，因为您使用的是IContextElem而不是IContext。

更改后，示例将在游乐场中再次编译。