当我直接调用函数时，这不是问题，但在这种情况下，通过另一个函数（如下面的示例）在内部调用时，Typescript无法捕获错误：

const updatePerson = ({
  name,
  onCurrentPerson,
}: {
  name: Person['name'];
  onCurrentPerson: (personValue: Person) => Person;
}) => {
  const updatedPeople = people.map((currentPerson) => {
    if (currentPerson.name !== name) return currentPerson;

    return onCurrentPerson(currentPerson);
  });
  return updatedPeople
};

请参见Typescript练习场上的以下代码

interface Person {
  name: string
  age: number
}

const people: Person[] = [{name: "John", age: 20}]

const updatePerson = ({
  name,
  onCurrentPerson,
}: {
  name: Person['name'];
  onCurrentPerson: (personValue: Person) => Person;
}) => {
  const updatedPeople = people.map((currentPerson) => {
    if (currentPerson.name !== name) return currentPerson;

    return onCurrentPerson(currentPerson);
  });
  return updatedPeople
};

updatePerson({
  name: "John",
  onCurrentPerson: (currentPerson) => {
    const person: Person = {
    ...currentPerson,
    age: 22,
    language: "English" // <-- Throws an error
    };
    return person;
  },
});

updatePerson({
  name: "John",
  onCurrentPerson: (currentPerson) => ({
    ...currentPerson,
    age: 22,
    language: "English"  // <-- Does not throw an error
    }),
});

正如您所看到的，我当前的解决方案要求我在返回之前检查值：

onCurrentPerson: (currentPerson) => {
  const person: Person = {
    ...currentPerson,
    age: 22,
    language: "English" // <-- Throws an error
  };
  return person;
},

我想了解为什么在我像下面的示例那样调用此错误时不显示它：

updatePerson({
  name: "John",
  onCurrentPerson: (currentPerson) => ({
    ...currentPerson,
    age: 22,
    language: "English"  // <-- Does not throw an error
  }),
});

我遗漏了什么？如何在返回值之前不必检查类型的情况下增强类型？