Intersection with a mapped type breaks bypassing private assignability.

🔎 Search Terms

private assignability intersection mapped type

🕗 Version & Regression Information

Latest as of today.

⏯ Playground Link

https://www.typescriptlang.org/play/?#code/CYUwxgNghgTiAEkoGdnwBrwN4CgCQADjAJYBuUALglAFzwB2ArgLYBGIMA3Pq3chSXoBzbgF8cOCgE8CCAJrwAvNni0GLdjHij4AMhW94-QUO3dps+AC0l8AArEwAawA8WVXSZsO2gDTwAcigAgD49Az4BYmEzCVAkOHgIEAp4AA86dG546ETk1Kk6OWzwXIR8+AAvOituHDTbKVU0BX0s+ttK5utw9qA

💻 Code

Given

declare class X {
	private a: number;
	b: string;
}

You're allowed to do:

type Y = { a: number } & { b: string };
declare let y: Y;
let x: X = y as Y & X;

But fails with:

type Z = Pick<{ a: number }, 'a'> & { b: string };
declare let z: Z;
let x: X = z as Z & X;

🙁 Actual behavior

X bypasses private assignability, but the identical type Y errors.

🙂 Expected behavior

Either neither type is allowed to bypass private assignability, or both types are, since they're identical.

Additional information about the issue

No response

从链接的 TypeScript Playground 复现开始，比较显式交集和基于 Pick 的交集的可赋值性。跟踪编译器对 private member 可赋值性的处理以及映射类型交集的行为，然后验证这两个等价示例产生一致的结果。