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即使我明确检查未定义,我也有一个简单的情况,strictNullChecks即不缩小类型。

interface WideType {
  one: { two: number } | undefined;
}
interface NarrowType {
  one: { two: number };
}

const example = (value: WideType): NarrowType | null => {
  if (value.one) {
    return value; // I have an error here
  }
  return null;
};

错误信息是:

Type 'WideType' is not assignable to type 'NarrowType'.
  Types of property 'one' are incompatible.
    Type '{ two: number; } | undefined' is not assignable to type '{ two: number; }'.
      Type 'undefined' is not assignable to type '{ two: number; }'.ts(2322)

我如何帮助 TS 编译器解决这个问题?TS 版本 3.7.2

4

2 回答 2

1

类型保护仅适用于类型,而不适用于属性,这意味着:

interface WideType {
  one: { two: number } | undefined;
}
interface NarrowType {
  one: { two: number };
}

const example = (value: WideType): NarrowType | null => {
  if (value.one) {
    // return value; // You have an error here, as value.one is not undefined, but value is still a WideType
    return {one: value.one}; // This works as value.one is not undefined
  }
  return null;
};
于 2019-11-15T09:14:05.923 回答
1

为了正确缩小类型,您可以在表单中创建自定义类型保护:

const isNarrow = (a: WideType): a is NarrowType => !!a.one;

在您的示例中使用:

interface WideType {
  one: { two: number } | undefined;
}
interface NarrowType {
  one: { two: number };
}

const isNarrow = (a: WideType): a is NarrowType => !!a.one;

const example = (value: WideType): NarrowType | null => {
  if (isNarrow(value)) {
    return value; // value is NarrowType
  }
  return null;
};
于 2019-11-15T14:09:49.343 回答