当没有类型注释时，Typescript 会提示有区别的联合类型

Question

在我看来， typescript 在没有任何类型注释的情况下无法识别可区分的联合类型。我错过了什么吗？有什么原因吗？

export type Stuff = AType | BType

export type AType = { status: Kind.A; name: string }

export type BType = { status: Kind.B; quantity: number }

export enum Kind {
  A,
  B,
}

function PlayWithStuff(stuff: Stuff) {
  console.log('some stuff', stuff)
}

const stuff = {
  status: Kind.B,
  quantity: 2,
}

PlayWithStuff(stuff)
//            ^^^^^
// Argument of type '{ status: Kind; quantity: number; }' is not assignable to parameter of type 'Stuff'.
//   Type '{ status: Kind; quantity: number; }' is not assignable to type 'BType'.
//     Types of property 'status' are incompatible.
//       Type 'Kind' is not assignable to type 'Kind.B'.

const secondStuff: Stuff = {
  status: Kind.B,
  quantity: 2,
}

PlayWithStuff(secondStuff)
// OK, the type annotation on `secondStuff` fixes the problem

Answer 1

初始化对象字面量时，Typescript 会推断属性类型，但不会推断为常量，因为它们不是只读的。
所以stuff的类型将是 { status: Kind, quantity: number } ，因为您以后可以将其更改为:

const stuff = {
    status: Kind.B,
    quantity: 2,
};

stuff.status = Kind.A;

所以现在它不能分配给 BType (也不是 AType 就此而言)。
您可以使用 as const :

const stuff = {
    status: Kind.B,
    quantity: 2,
} as const;

现在类型推断为 { readonly status: Kind.B, readonly quantity: 2}它始终可分配给 BType。
或者你可以做你所做的，只是给它类型注释:

const stuff: BType = {
    status: Kind.B,
    quantity: 2,
};

stuff.status = Kind.A; // Errors