Scala如何以不同方式推断这两个函数？

Question

这编译:

 def fibonacci():() => Int = {
        var first   = 1
        var second  = 2
        return () => {
            val oldFirst = first
            first = second
            second = second + oldFirst
            second 
        }
    }

这不是:

 def fibonacci():() => Int = {
     var first   = 1
     var second  = 2
     return ():Int => {
         val oldFirst = first
         first = second
         second = second + oldFirst
         second 
     }
 }

我明确地试图告诉它我正在返回一个Int,但是我得到了这个错误:Illegal start of declaration并且它指向该first = second行.他们有什么不同？我正在使用Scala 2.8.1.

Answer 1

return (): Int => {...}在Scala中不是一个恰当的表达.如果要显式指定返回类型,则需要在值之后放置声明(并且值将是匿名函数):

def fibonacci():() => Int = {
  var first   = 1
  var second  = 2
  return ( () => {
    val oldFirst = first
    first = second
    second = second + oldFirst
    second 
  } ) : () => Int
}

但请注意,没有必要这样做.如果省略return,则根本不需要进行任何显式类型声明:

def fibonacci() = {
  var first   = 1
  var second  = 2
  () => {
    val oldFirst = first
    first = second
    second = second + oldFirst
    second 
  }
}

Answer 2

Debilski是对的.还有两条评论:(1)return关键字不是必需的.默认情况下,最后一个表达式成为返回值.(2)如果你试图单独注释函数体的类型,那是可能的.代码变成:

def fibonacci2(): () => Int = {
  var first   = 1
  var second  = 2
  () => {
    val oldFirst = first
    first = second
    second = second + oldFirst
    second 
  }: Int
}