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typescript - 键入复合类型保护 - 从类型保护推断/继承属性到其调用者

问题描述

type Bird = { fly: () => "fly" };
type Insect = { annoy: () => "annoy" };
type Dog = { beTheBest: () => "dogdogdog" };

type Animal = Bird | Insect | Dog;

const isBird = (animal: Animal): animal is Bird => {
  if ("fly" in animal) return true;
  return false;
};

const isInsect = (animal: Animal): animal is Insect => {
  if ("annoy" in animal) return true;
  return false;
};

const hasWings = (animal: Animal) => {
  if (isBird(animal)) return true;
  if (isInsect(animal)) return true;
  return false;
};

我将两个基本的类型保护功能组合到复合hasWings保护中，但是 TypeScript 并没有推断出它的类型保护特性——它typeof hasWings只是简单地推断出(a: Animal) => boolean. 有没有一种方法可以帮助 TS 推断或明确告诉它hasWings是isInsect和isBird类型保护的组合，而无需手动重新指定标准？

// Something like this would be useful to me:
const hasWings = (animal: Animal): ReturnType<typeof isBird> & ReturnType<typeof isInsect> => {
  if (isBird(animal)) return true;
  if (isInsect(animal)) return true;
  return false;
};

// Having to specify the whole list manually is not useful to me:
const hasWings = (animal: Animal): animal is Insect | Bird => {
  if (isBird(animal)) return true;
  if (isInsect(animal)) return true;
  return false;
};

标签： typescripttypeguards

解决方案

您可以编写一个辅助函数来以这种方式组合类型保护。类型注释很拗口，但它有效：

type TypeGuard<S, T extends S> = (x: S) => x is T;

function typeGuardUnion<T extends TypeGuard<any, any>[]>(...fs: T): TypeGuard<
    T extends TypeGuard<infer S, any>[] ? S : never,
    T extends TypeGuard<any, infer U>[] ? U : never
> {
    return ((x: any) => fs.some(f => f(x))) as any;
}

// TypeGuard<Animal, Bird | Insect>
const hasWings = typeGuardUnion(isBird, isInsect);

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