TypeScript 当枚举有>1个成员时，仅针对枚举进行穷举性检查,

7cwmlq89 于 9个月前发布在 TypeScript

关注(0)|答案(5)|浏览(143)

TypeScript版本： typescript@2.9.0-dev.20180420
搜索词： 区分，穷举，类型保护，缩小
代码

// Legal action types for ValidAction
enum ActionTypes {
  INCREMENT = 'INCREMENT',
//   DECREMENT = 'DECREMENT',
}
interface IIncrement {
  payload: {};
  type: ActionTypes.INCREMENT;
}
// interface IDecrement {
//   payload: {};
//   type: ActionTypes.DECREMENT;
// }
// Any string not present in T
type AnyStringExcept<T extends string> = { [P in T]: never; };
// ValidAction is an interface with a type in ActionTypes
type ValidAction = IIncrement;
// type ValidAction = IIncrement | IDecrement;
// UnhandledAction in an interface with a type that is not within ActionTypes
type UnhandledAction = { type: AnyStringExcept<ActionTypes>; };
// The set of all actions
type PossibleAction = ValidAction | UnhandledAction;
// Discriminates to ValidAction
function isUnhandled(x: PossibleAction): x is UnhandledAction {
    return !(x.type in ActionTypes);
}
type CounterState = number;
const initialState: CounterState = 0;
function receiveAction(state = initialState, action: PossibleAction) {
    // typeof action === PossibleAction
    if (isUnhandled(action)) {
        // typeof action === UnhandledAction
        return state;
    }
    // typeof action === ValidAction
    switch (action.type) {
        case ActionTypes.INCREMENT:
            // typeof action === IIncrement
            return state + 1;
        // case ActionTypes.DECREMENT:
        //     return state - 1;
    }
    // typeof action === IIncrement
    // Since INCREMENT is handled above, this should be impossible,
    // However the compiler will say that assertNever cannot receive an argument of type IIncrement
    return assertNever(action);
}
function assertNever(x: UnhandledAction): never {
    throw new Error(`Unhandled action type: ${x.type}`);
}

预期行为： 不会抛出错误，因为switch语句是穷举的。如果取消注解ActionTypes.DECREMENT部分(导致ActionTypes有两个可能的值),则不会出现错误。只有在ActionTypes只有一个值时才会出现错误。即使在默认情况下发生neverAssert，错误仍然会发生，这显然是从IIncrement不可达的。
实际行为： 尽管只有可能的值被显式处理，但仍然抛出错误。如果取消注解ActionTypes.DECREMENT,预期行为将出现。
Playground链接： (修复了链接)

错误
工作

相关问题：

#19904
#14210
#18056

typescript

来源：https://github.com/microsoft/TypeScript/issues/23572

5条答案

按热度按时间

jvidinwx1#

这也是同样的情况，当你尝试为未来的联合类型准备代码，但目前只使用一种类型。
如果你取消注解所有与第二个接口(前3个注解块)相关的内容，一切都将按预期工作。

赞(0）回复(0）举报 9个月前

daolsyd02#

是的，这确实只是归结为"当对大小大于1的工会进行区分时，穷尽性检查才能起作用"。

赞(0）回复(0）举报 9个月前

o3imoua43#

我刚刚也遇到了这个bug。这里有一个简单的例子：

function assertNever(x: never): never {
  throw new Error("Unexpected object: " + x);
}
interface Square {
  kind: "square";
  size: number;
}
type Shape = Square
function area(s: Shape) {
  switch (s.kind) {
  case "square": return s.size * s.size;
  default:
    assertNever(s);
  }
}

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赞(0）回复(0）举报 9个月前

xtfmy6hx4#

我们今天也遇到了这个问题。这里有另一个代码示例，如果它能帮助的话

const rejectUnexpectedValueOfPropertyInObject = (
  objectName: string,
  propertyName: string,
  object: never
): never => {
  throw new Error(
    `Unexpected value for ${propertyName} in ${objectName}: ${
object && typeof object === "object"
? (object as any)[propertyName]
: "<not an object>"
}`
  );
};
type Result =
  | {
      outcome: "success";
    }
  // | {
  //    outcome: "error";
  //    reason: "reason_2";
  //  }
  | {
      outcome: "error";
      reason: "reason_1";
    };
const f = (a: number, b: number): Result => {
  if (a < 0) {
    return {
      outcome: "error",
      reason: "reason_1"
    };
  }
  return { outcome: "success" };
};
const run = (a: number, b: number) => {
  const result = f(1, 2);
  if (result.outcome === "error") {
    switch (result.reason) {
      case "reason_1":
        console.log("error reason 1");
        break;
      // case "reason_2":
      //  console.log("error reason 2");
      //  break;
      default:
        rejectUnexpectedValueOfPropertyInObject("result", "reason", result);
    }
  }
};

错误是

const result: {
            outcome: "error";
            reason: "reason_1";
          }
          Argument of type '{ outcome: "error"; reason: "reason_1"; }' is not assignable to parameter of type 'never'.

通过移除类型定义和switch中的reason_2的注解，错误消失了。

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赞(0）回复(0）举报 9个月前

moiiocjp5#

今天在尝试为扩展结构化一些代码时遇到了这个问题：
https://www.typescriptlang.org/play/?ssl=22&ssc=2&pln=1&pc=1#code/JYOwLgpgTgZghgYwgAgJIgM7TAFQgDzGQG8AoZC5MATwAcIAuZAclCyjAH1JDmBucpVpwwACyYgArgFsARtAGUqBMEwxgooAOYCAvqVI16yAPL0oI4AHsQyALxpM2PIQGkYkkAjDXbETJJQEACqIEGIonCyADYQABQAbnDRkozIIBAJ0AA0yNIQGBhwWhAA-Goa2gCUEpnQJIJiUFYA7ukQbQCiUM1QcfmFxSgAPsPIAAah4QiRMSgIVgAmKKLARC1roshJKSgAJMQAypUgWonJqVW641V6Bh5ePjbIcLS00dRmcVa0TGbQlhsNW2VmAiwaSgwGzAM2Q31oADojBAqiRkIIlBQEHAsCw2NhuCpmEwyJiycgFpgrLEEdErGcfkiVLcMeSKEEwIEQIpyfo2ct4JJoqpWWz-BhAiEwhAIlFYvCWUp9LogA

赞(0）回复(0）举报 9个月前

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TypeScript 当枚举有>1个成员时，仅针对枚举进行穷举性检查,

5条答案

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