为什么Option.fold的结果是一个函数？

当评估Option的数据时,我试图使用fold[B](ifEmpty: => B)(f: A => B): B,但我对结果有点困惑:

scala> Some(1).fold(() => "empty")(d => d.toString)
res5: () => String = 

scala> Some(1).fold(() => "empty")(d => d.toString)()
res6: String = 1

这个方法的源代码:

  @inline final def fold[B](ifEmpty: => B)(f: A => B): B =
    if (isEmpty) ifEmpty else f(this.get)

我预计结果是String,但得到了,为什么？

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我尝试使用以下代码模拟这种情况:

case class Demo(size: Int)

// version 1, with the same fold method as Option
case class X [A <: Demo](data: A) {
    def fold[B](ifEmpty: => B)(f: A => B): B = {
        if(data.size < 3) ifEmpty else f(data)
    }
}

val demo = Demo(2)
val x = X(demo)

x.fold(() => "empty")(d => d.toString)   // the result is a function

// version 2
case class X [A <: Demo](data: A) {
    def fold[B](g: A => B)(f: A => B): B = {
        if(data.size < 3) g(data) else f(data)
    }
}

x.fold(g => g.toString)(f => f.toString)   // the result is a String

根据演示,似乎结果受到scala类型推断的影响,是吗？