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

Inizia con la riproduzione collegata in TypeScript Playground e confronta l’assegnabilità per l’intersezione esplicita e l’intersezione basata su Pick. Traccia la gestione da parte del compilatore dell’assegnabilità dei membri privati e il comportamento delle intersezioni di tipi mappati, quindi verifica che i due esempi equivalenti producano risultati coerenti.