Question

我有一个递归函数，它将重复该函数，直到不满足if条件，然后输出一个整数。但是，此函数外部需要整数的函数正在接收一个单元。我应该如何修改代码以返回int？

count(r,c,1,0)

   def count(r: Int, c: Int, countR: Int, lalaCount: Int): Int = {
    if (countR < (r + 1)) count(r,c,countR + 1, lalaCount + countR)
    else (lalaCount + c + 1)
   }

这是整个计划

object hw1 {
  def pascal(c: Int, r: Int): Int = {

   count(r,c,1,0)

   def count(r: Int, c: Int, countR: Int, lalaCount: Int): Int = {
    if (countR < (r + 1)) count(r,c,countR + 1, lalaCount + countR)
    else (lalaCount + c + 1)
   }
  } //On this line eclipse is saying "Multiple markers at this line
    //- type mismatch;  found   : Unit  required: Int
    //- type mismatch;  found   : Unit  required: Int
pascal(3,4)

}

Answer 1

pascal返回的值是它包含的最后一个表达式。你希望它是你对count的评价，但这不是最后一件事。分配（def，val等）的类型为Unit，如您所见：

  def pascal(c: Int, r: Int): Int = {

   count(r,c,1,0) // => Int

   def count(r: Int, c: Int, countR: Int, lalaCount: Int): Int = {
    if (countR < (r + 1)) count(r,c,countR + 1, lalaCount + countR)
    else (lalaCount + c + 1)
   } // => Unit
  }

在 count(r,c,1,0)之后移动def ，这样可以解决问题。

递归函数不返回Int

1 个答案: