Question

我想编写一个函数，该函数返回扩展的泛型类型的属性。甚至有可能吗？

这是我尝试过的：

interface Animal {
    readonly weight: {total: number}
}

interface Dog extends Animal {
    readonly weight: {total: number, tail: number, head: number, body: number }
}


const getWeight = <T extends Animal>(animal: T) => {
    return animal.weight
}

const myDog = { weight: { total: 20, tail: 5, head: 5, body: 10 } }

const myDogsWeight = getWeight<Dog>(myDog)

// Property 'head' does not exist on type '{ total: number; }'.
const myDogsHeadWeight = myDogsWeight.head

如果尝试显式注释函数的返回类型，则会收到另一个错误：

type InferredWeight<TAnimal> = TAnimal extends { readonly weight: infer U } ? U : never
type DogWeight = InferredWeight<Dog> // <-- This type works correctly

const getWeightTyped = <T extends Animal>(animal: T): InferredWeight<T> => {
    // Type '{ total: number; }' is not assignable to type 'InferredWeight<T>'.
    return animal.weight
}

这是playground link。

看起来唯一相关的TypeScript Github问题是this one，但这是联合类型的问题。

Answer 1

如@jcalz在评论中所述，您可以使用查找类型：

这应该可以满足您的需求。

interface Animal {
  readonly weight: { total: number }
}

interface Dog extends Animal {
  readonly weight: { total: number, tail: number, head: number, body: number }
}

const getWeight = <
  T extends Animal
>(animal: T): T['weight'] => {
  return animal.weight
}

const myDog = { weight: { total: 20, tail: 5, head: 5, body: 10 } }

const myDogsWeight = getWeight(myDog);

如何推断扩展泛型类型的子属性类型？

1 个答案: