Question

我有一个简单的情况，即使我显式检查未定义，strictNullChecks也不缩小类型。

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;
};

错误消息是：

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编译器解决这一问题？ TS版本3.7.2

Answer 1

类型防护仅适用于类型，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;
};

Answer 2

为了适当地缩小类型，您可以采用以下形式创建自定义类型防护：

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;
};

打字稿strictNullChecks不缩小类型

2 个答案: