在函数类型中定义返回类型时打字稿额外的返回属性

Question

The following example illustrates my issue pretty well:

type Position = {
  x: number
  y: number
}

const update = (): Position => ({ x: 0, y: 0, z: 0 }) // 'z' does not exist in type 'Position', as it should.

type Update = () => Position

const update: Update = () => ({ x: 0, y: 0, z: 0 }) // that's fine?

const position: Position = update() // so is this, even though it is clearly wrong. 
                            

What if I want to type this functions in a separate file? Other than adding the return type to the actual function.

Answer 1

When you declare that an object must be of an interface, you are saying that it must have at least those properties, not only those properties.

Typescript will complain when you are creating an object from scratch (aka an object literal) with properties that are not known to the interface, but if the object is a variable then you won't get any error when using it to fulfill an interface.

This is ok:

const xyz = { x: 0, y: 0, z: 0 };

const position1: Position = xyz;

But this gives error "Object literal may only specify known properties, and 'z' does not exist in type 'Position'"

const position2: Position = { x: 0, y: 0, z: 0 };

Your final example const position: Position = update() is actually fine because update() returns something that extends Position . You could be even more restrictive and say const position: {x: number} = update() and that's fine too, as long at the value has at least {x: number} . The object in your variable position will still have y and z properties, but you'll get an error when trying to access them because typescript doesn't know about them.