Typescript 类型推断、传播语法和多种类型 return

Question

interface SkillProperty {
    [name: string] : number 
};

let skills: SkillProperty;

skills = {}; // ok

skills = { fire: 123 }; // ok

skills = {
    ...skills, // ok
    ...{}, // ok
    ...extraSkills() // {} | { ice: number } is not assignable to type 'SkillProperty'.
}

function extraSkills() {
    if (whatever) {
        return {};
    }
    return { ice: 321 };
}

如何更改我的 SkillProperty 接口以使其与空对象和我的实际 SkillProperty 类型兼容？

Answer 1

您的 SkillProperty 接口实际上与 {} | {ice: number}:

兼容

let noSkills = {}
let iceSkills = { ice: 321 };
let randomSkills: {} | {ice: number} = (Math.random() < 0.5) ? noSkills : iceSkills
let maybeSkills: SkillProperty = randomSkills; // no error

所以，对我来说，这看起来像是 TypeScript 中的一个错误。类似的 issue was fixed, but this case 似乎仍然存在。可能值得打开一个包含现有问题链接的新问题。

同时有变通办法，例如：

skills = {
  ...skills, // ok
  ...{}, // ok
  ...extraSkills() as SkillProperty // okay now
}

希望对您有所帮助；祝你好运！

Typescript 类型推断、传播语法和多种类型 return

Typescript type inference, spread syntax and multiple type return

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