scalaz 验证和列表 monad

发布于 2024-11-28 21:05:17 字数 300 浏览 1 评论 0原文

我试图提出类似于以下内容的内容：

val s: Validation[String, Int] = 1.success
def s2(i: Int): Validation[String, Int] = i.success

val result = for {
  i <- s
  j <- List(1, 2)
  k <- s2(j)
} yield "fine";

上面的代码无法编译，我理解，从语法上讲它没有意义。

我正在尝试以单子方式执行验证列表。我该如何实现这一目标？

原文

I am trying to come up with something similar to the following:

val s: Validation[String, Int] = 1.success
def s2(i: Int): Validation[String, Int] = i.success

val result = for {
  i <- s
  j <- List(1, 2)
  k <- s2(j)
} yield "fine";

The above code does not compile and I understand, syntactically it does not make sense.

I am trying to execute a list of validations in a monadic way. How do I achieve that?

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上课铃就是安魂曲 2024-12-05 21:05:17

如果您有 A 的验证列表，则可以使用 sequence 将其转换为 A 列表的验证：（

List(1, 2).map(s2).sequence[({type l[a]=Validation[String, a]})#l, Int]

如果我理解正确提问）。所以你得到

val result = for {
  i <- s
  k <- List(1, 2).map(s2).sequence[({type l[a]=Validation[String, a]})#l, Int] 
} yield "fine"

If you have a list of validations of A, you can turn it into a validation of lists of A using sequence:

List(1, 2).map(s2).sequence[({type l[a]=Validation[String, a]})#l, Int]

(if I understand the question correctly). So you get

val result = for {
  i <- s
  k <- List(1, 2).map(s2).sequence[({type l[a]=Validation[String, a]})#l, Int] 
} yield "fine"

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下雨或天晴 2024-12-05 21:05:17

您似乎正在使用验证来消除副作用。这不是它的目的。您可以在函数式编程中使用返回值。

for 理解中的验证在成功时继续，但在失败时中断并返回失败。

scala> def g(i: Int): Validation[String, Int] = { 
          println(i); if(i % 2 == 0) i.success else "odd".fail 
       }
g: (i: Int)scalaz.Validation[String,Int]

scala> val result = for {
     |   i <- g(1)
     |   j <- g(2)
     | } yield (i,j)
1
result: scalaz.Validation[String,(Int, Int)] = Failure(odd)

scala> val result = for {
     |   i <- g(2)
     |   j <- g(1)
     | } yield (i,j)
2
1
result: scalaz.Validation[String,(Int, Int)] = Failure(odd)


scala> val result = for {
     |   i <- g(2)
     |   j <- g(2)
     | } yield (i,j)
2
2
result: scalaz.Validation[String,(Int, Int)] = Success((2,2))


scala> val result = for {
     |   i <- g(1)
     |   j <- g(1)
     | } yield (i,j)
1
result: scalaz.Validation[String,(Int, Int)] = Failure(odd)

You seem to be using validation for the side effect. This is not what its ment for. You use the return values in functional programming.

Validation in a for comprehension continues with on success, but breaks of at a failure and returns the failure.

scala> def g(i: Int): Validation[String, Int] = { 
          println(i); if(i % 2 == 0) i.success else "odd".fail 
       }
g: (i: Int)scalaz.Validation[String,Int]

scala> val result = for {
     |   i <- g(1)
     |   j <- g(2)
     | } yield (i,j)
1
result: scalaz.Validation[String,(Int, Int)] = Failure(odd)

scala> val result = for {
     |   i <- g(2)
     |   j <- g(1)
     | } yield (i,j)
2
1
result: scalaz.Validation[String,(Int, Int)] = Failure(odd)


scala> val result = for {
     |   i <- g(2)
     |   j <- g(2)
     | } yield (i,j)
2
2
result: scalaz.Validation[String,(Int, Int)] = Success((2,2))


scala> val result = for {
     |   i <- g(1)
     |   j <- g(1)
     | } yield (i,j)
1
result: scalaz.Validation[String,(Int, Int)] = Failure(odd)

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