为什么我不能返回一个通用的 'T' 来满足 Partial<t> ?</t>

Question

I wrote some code in TypeScript:

type Point = {
  x: number;
  y: number;
};
function getThing<T extends Point>(p: T): Partial<T> {
  // More interesting code elided
  return { x: 10 };
}

This produces an error:

Type '{ x: 10; }' is not assignable to type 'Partial<T>'

This seems like a bug - { x: 10 } is clearly a Partial<Point> . What's TypeScript doing wrong here? How do I fix this?

Answer 1

When thinking about writing a generic function, there's an important rule to remember

The Caller Chooses the Type Parameter

The contract you've provided for getThing ...

function getThing<T extends Point>(p: T): Partial<T>

... implies legal invocations like this one, where T is a subtype of Point :

const p: Partial<Point3D> = getThing<Point3D>({x: 1, y: 2, z: 3});

Of course, { x: 10 } is a legal Partial<Point3D> .

But the ability to subtype doesn't just apply to adding additional properties -- subtyping can include choosing a more restricted set of the domain of the properties themselves. You might have a type like this:

type UnitPoint = { x: 0 | 1, y: 0 | 1 };

Now when you write

const p: UnitPoint = getThing<UnitPoint>({ x: 0, y: 1});

px has the value 10 , which is not a legal UnitPoint .

If you find yourself in a situation like this, odds are good that your return type is not actually generic . A more accurate function signature would be

function getThing<T extends Point>(p: T): Partial<Point> {