Scala理解中的println [英] println in scala for-comprehension

问题描述

出于理解，我不能只发表一份印刷声明:

In a for-comprehension, I can't just put a print statement:

def prod (m: Int) = {
  for (a <- 2 to m/(2*3);
    print (a + "  ");
    b <- (a+1) to m/a;
    c = (a*b) 
    if (c < m)) yield c
}

但是我可以通过一个虚拟任务轻松地绕开它:

but I can circumvent it easily with a dummy assignment:

def prod (m: Int) = {
  for (a <- 2 to m/(2*3);
    dummy = print (a + "  ");
    b <- (a+1) to m/a;
    c = (a*b) 
    if (c < m)) yield c
}

有副作用，并且仅在开发中的代码中使用(到目前为止)，有没有更好的临时解决方案?

Being a side effect, and only used (so far) in code under development, is there a better ad hoc solution?

除了有副作用之外，为什么我不应该使用它还有一个严重的问题吗?

Is there a serious problem why I shouldn't use it, beside being a side effect?

从与雷克斯·克尔(Rex Kerr)的讨论开始，有必要显示原始代码，这有点复杂，但似乎与问题无关(2x .filter，最后调用一个方法)，但是当我尝试将Rex模式应用到它时，我失败了，所以我将其发布在这里:

From the discussion with Rex Kerr, the necessity has risen to show the original code, which is a bit more complicated, but did not seem to be relevant for the question (2x .filter, calling a method in the end), but when I tried to apply Rex' pattern to it I failed, so I post it here:

  def prod (p: Array[Boolean], max: Int) = {
    for (a <- (2 to max/(2*3)).
        filter (p);
      dummy = print (a + "  ");
      b <- (((a+1) to max/a).
         filter (p));
      if (a*b <= max)) 
        yield (em (a, b, max)) }

这是我的尝试-(b * a).filter错误，因为结果是一个int，而不是可过滤的int集合:

Here is my attempt -- (b * a).filter is wrong, because the result is an int, not a filterable collection of ints:

  // wrong: 
  def prod (p: Array[Boolean], max: Int) = {
    (2 to max/(2*3)).filter (p).flatMap { a =>
      print (a + " ")
      ((a+1) to max/a).filter (p). map { b => 
        (b * a).filter (_ <= max).map (em (a, b, max))
      }
    }
  }

第二部分属于注释，但如果在此处写，则无法阅读-也许我最后将其删除.请原谅.

Part II belongs to the comments, but can't be read, if written there - maybe I delete it in the end. Please excuse.

好-这是Rex在代码布局中的最后答案:

Ok - here is Rex last answer in code layout:

  def prod (p: Array[Boolean], max: Int) = {
    (2 to max/(2*3)).filter (p).flatMap { a =>
      print (a + " ")
      ((a+1) to max/a).filter (b => p (b) 
        && b * a < max).map { b => (m (a, b, max))
      }
    }
  }

推荐答案

这是您需要编写的方式:

This is how you need to write it:

scala> def prod(m: Int) = {
     |   for {
     |     a <- 2 to m / (2 * 3)
     |     _ = print(a + " ")
     |     b <- (a + 1) to (m / a)
     |     c = a * b
     |     if c < m
     |   } yield c
     | }
prod: (m: Int)scala.collection.immutable.IndexedSeq[Int]

scala> prod(20)
2 3 res159: scala.collection.immutable.IndexedSeq[Int] = Vector(6, 8, 10, 12, 14
, 16, 18, 12, 15, 18)

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