typescript - 如何告诉 Typescript 从一个不太严格的接口(interface)转换到一个更严格的接口(interface)是可以的？

Question

我试图让我粘贴在下面的代码执行，但 Typescript 编译器似乎不相信我已经正确地确保一个接口(interface)可以根据我所做的检查转换为另一个接口(interface)。我需要做什么才能让它对这段代码满意？

interface Person {
  name: string,
  address?: string
}

interface PersonWithAddress extends Person {
  address: string
}

function tryLoggingAddresses(people: Array<Person>) {
  people.forEach(person => {
    if (person.address) {
      logAddress(person)
    }
  })
}

function logAddress(personWithAddress: PersonWithAddress) {
  console.log(personWithAddress.address)
}

typescript 错误:

error TS2345: Argument of type 'Person' is not assignable to parameter of type 'PersonWithAddress'.
      Types of property 'address' are incompatible.
        Type 'string | undefined' is not assignable to type 'string'.
          Type 'undefined' is not assignable to type 'string'.

    63       logAddress(person)

我知道我可以通过将行更改为 logAddress(person as PersonWithAddress) 来强制转换，但这实际上禁用了所有不理想的类型检查。如果没有那个，有没有办法做到这一点？

Answer 1

您可以使用 type guard代替您现有的条件表达式。这将有助于通知编译器数据结构满足您在 control flow analysis 期间想要的标准。 .

TypeScript Playground

function personHasAddress (person: Person): person is PersonWithAddress {
  return typeof person.address === 'string';
}

function logAddresses (people: Person[]) {
  for (const person of people) {
    if (personHasAddress(person)) logAddress(person);
  }
}