为什么函数重载可以解决类型缩小问题？

Question

我遇到了一个问题，它基本上涉及缩小从一个对象中编辑的类型 return，通过它的键访问。假设我有一个对象，其值可以是 number 或 string:

type Value = number | string;

enum DatumKey {
  numberDatum = 'numberDatum',
  stringDatum = 'stringDatum',
}

class DemoClass {
  private data = {
    [DatumKey.numberDatum]: 1337,
    [DatumKey.stringDatum]: 'foobar',
  }

  public getValue(key: DatumKey): Value {
    return this.data[key];
  }

  public doSomething(): void {
    const num: number = this.getValue(DatumKey.numberDatum); // TS expects `number | string`, I expect `number`
    const str: string = this.getValue(DatumKey.stringDatum); // TS expects `number | string`, I expect `string`

    console.log({ num, str });
  }
}

const dc = new DemoClass();
dc.doSomething();

See Typescript Playground example.

方法失败

所以，为了缩小类型，我尝试使用通用类型，然后我可以告诉 getValue<...>() 我在 return 中期望什么样的类型：

public getValue<T extends Value>(key: DatumKey): T {
  // Error here: Type 'Value' not assignable to type 'T'
  return this.data[key];
}

public doSomething(): void {
  const num: number = this.getValue<number>(DatumKey.numberDatum); // TS expects `number`, it's good!
  const str: string = this.getValue<string>(DatumKey.stringDatum); // TS expects `string`, it's good!
}

但是，通过这样做，我从 TypeScript 中得到一个错误：

Type 'Value' is not assignable to type 'T'.
  'Value' is assignable to the constraint of type 'T', but 'T' could be instantiated with a different subtype of constraint '{}'.
    Type 'string' is not assignable to type 'T'.
      'string' is assignable to the constraint of type 'T', but 'T' could be instantiated with a different subtype of constraint '{}'.

See Typescript Playground example.

工作方式

然后我在这里遇到了一个类似的问题，。确实如此。但是，我不确定为什么函数重载可以解决问题：

public getValue<T extends Value>(key: DatumKey): T 
public getValue(key: DatumKey): Value {
  return this.data[key];
}

public doSomething(): void {
  const num = this.getValue<number>(DatumKey.numberDatum); // TS expects `number`, it's good!
  const str = this.getValue<string>(DatumKey.stringDatum); // TS expects `string`, it's good!
}

See Typescript Playground example.

Answer 1

您的工作解决方案并不真正有效。变量 num 和 str 的类型分别为 Value 而不是 number 和 string.

更强大的解决方案：

enum DatumKey {
  numberDatum = 'numberDatum',
  stringDatum = 'stringDatum',
}

const data = {
  [DatumKey.numberDatum]: 1337,
  [DatumKey.stringDatum]: 'foobar',
};

function getValue<T extends keyof typeof data>(key: T): (typeof data)[T] {
  return data[key];
}

// You don't have to explicitly type num and str. They are automatically inferred 
const num = getValue(DatumKey.numberDatum);
const str = getValue(DatumKey.stringDatum);

为什么函数重载可以解决类型缩小问题？

Why does function overload fix type-narrowing issue?

typescript

javascript-objects

type-narrowing

方法失败

工作方式