Question

我试图根据选项值返回Either值。我的目标是，如果存在该选项，则返回Either.right（），否则代码应返回Either.left（）。我使用Java 8和vavr 0.9.2

我想避免条件性禁忌

public Either<String, Integer> doSomething() {
    Optional<Integer> optionalInteger = Optional.of(Integer.MIN_VALUE);
    Option<Integer> integerOption = Option.ofOptional(optionalInteger);

    return integerOption.map(value -> {
      //some other actions here
      return Either.right(value);
    }).orElse(() -> {
      //some other checks her also 
      return Either.left("Error message");
    });
}

编译器失败，并显示以下消息

Error:(58, 7) java: no suitable method found for orElse(()->Either[...]age"))
    method io.vavr.control.Option.orElse(io.vavr.control.Option<? extends io.vavr.control.Either<java.lang.Object,java.lang.Integer>>) is not applicable
      (argument mismatch; io.vavr.control.Option is not a functional interface
          multiple non-overriding abstract methods found in interface io.vavr.control.Option)
    method io.vavr.control.Option.orElse(java.util.function.Supplier<? extends io.vavr.control.Option<? extends io.vavr.control.Either<java.lang.Object,java.lang.Integer>>>) is not applicable
      (argument mismatch; bad return type in lambda expression
          no instance(s) of type variable(s) L,R exist so that io.vavr.control.Either<L,R> conforms to io.vavr.control.Option<? extends io.vavr.control.Either<java.lang.Object,java.lang.Integer>>)

Answer 1

orElse返回Option<T>，而doSomething返回类型要求Either<String, Integer>。

相反，请尝试使用返回getOrElse的{{1}}：

Answer 2

由于您要返回PLtz80rG4U0K99mMxPLNFPhwBIKkvzx9x1而不是Either<...>，因此必须使用uploaded videos page from HRVY channel

Answer 3

我认为@ oleksandr-pyrohov的答案中多余的泛型和格式会使事情变得有些复杂。我会这样写：

public Either<String, Integer> doSomething() {
    // ...
    return integerOption
        .map(Either::right)
        .orElseGet(() -> Either.left("Error message"));
}

更新。：尽管似乎并不总是可悲地推断出泛型。

如何根据Option值向左/向右获取

3 个答案: