Question

我有以下验证逻辑：

  def one(a : String) : Validation[String, Int] =
    if (a == "one") {
      Success(1)
    } else {
      Failure("Not One")
    }

  def two(a : String) : Validation[String, Int] =
    if (a == "two") {
      Success(2)
    } else {
      Failure("Not Two")
    }

  def validate (a : String) = (one(a) |@| two(a)){_  + _}

根据Scalaz文档：

  /**
   * DSL for constructing Applicative expressions.
   *
   * `(f1 |@| f2 |@| ... |@| fn)((v1, v2, ... vn) => ...)` is an alternative to `Apply[F].applyN(f1, f2, ..., fn)((v1, v2, ... vn) => ...)`
   *
   * `(f1 |@| f2 |@| ... |@| fn).tupled` is an alternative to `Apply[F].applyN(f1, f2, ..., fn)(TupleN.apply _)`
   *
   * Warning: each call to `|@|` leads to an allocation of wrapper object. For performance sensitive code, consider using
   *          [[scalaz.Apply]]`#applyN` directly.
   */

如何将验证功能转换为使用apply2？

Answer 1

Validate的类型构造函数有两个参数，但Apply只能由arity one的类型构造函数参数化。你需要一个名为lambda的特殊技巧，它允许我们讨论类型定义：

def validate(a : String) = Apply[({type λ[Int] = Validation[String, Int]})#λ].apply2(one(a), two(a)){_  + _}

如何使用申请功能应用程序

1 个答案: