switch 语句中的 Typescript 类型生成问题

Question

我有这个映射函数，它正在渲染一个对象的类型。

function mapStuffs(fields: any) {
    switch (fields.type) {
        case "NewStuffs":
            return {
                type: "NewStuffs" as const,
                stuff: fields.stuff
            };
        case "StuffFile":
            const fileName = "jadflkjs";
            const stuffId = "adlkajsf";
            return {
                type: "StuffFile" as const,
                fileName,
                stuffId
            };
        default:
            return {
                type: "ExistingStuff" as const
            };
    }
}

当我构建时，生成的类型是:

mainstuff:
        | {
              stuff: string | undefined;
              type: "NewStuffs";
              fileName?: undefined;
              stuffId?: undefined;
          }
        | {
              type: "StuffFile";
              fileName: string;
              stuffId: string;
          }
        | {
              type: "ExistingStuff";
              fileName?: undefined;
              stuffId?: undefined;
          };

我不明白如何删除不需要的 fileName 和 stuffId 作为可选参数。这是 typescript 错误吗？我想要的是这种类型:

mainstuff:
        | {
              stuff: string | undefined;
              type: "NewStuffs";
          }
        | {
              type: "StuffFile";
              fileName: string;
              stuffId: string;
          }
        | {
              type: "ExistingStuff";
          };

Answer 1

这不是 TypeScript 错误。这是自 TypeScript 2.7 引入以来的预期行为 improved type inference for object literals . fileName 和 stuffId 未包含在某些返回值中这一事实使编译器认为缺失应该反射(reflect)在返回类型中。

TypeScript 中的类型推断算法是一组启发式算法，语言设计者试图猜测在特定情况下大多数人希望看到什么样的东西。如果你不喜欢推断的类型，你可以随意 annotate the return type您的 mapStuffs() 函数与您喜欢的任何类型，编译器将对此感到满意，假设您的函数实现符合它:

type MappedStuff = {
    stuff: string | undefined;
    type: "NewStuffs";
} | {
    type: "StuffFile";
    fileName: string;
    stuffId: string;
} | {
    type: "ExistingStuff";
};

function mapStuffs(fields: any): MappedStuff {

    switch (fields.type) {
        case "NewStuffs":
            return {
                type: "NewStuffs" as const,
                stuff: fields.stuff
            };
        case "StuffFile":
            const fileName = "jadflkjs";
            const stuffId = "adlkajsf";
            return {
                type: "StuffFile" as const,
                fileName,
                stuffId
            };
        default:
            return {
                type: "ExistingStuff" as const
            };
    }
}

Playground link to code