使用 Scala 流跟踪斐波那契计算的执行

Question

我是函数式编程/scala新手。我一直在尝试让我的头脑围绕以下代码片段和生成的输出。

def fib:Stream[Int] = {
  Stream.cons(1,
    Stream.cons(2,
      (fib zip fib.tail) map {case (x, y) => println("%s + %s".format(x, y)); x + y}))
}

输出跟踪：

scala> fib take 4 foreach println
1
2
1 + 2
3
1 + 2  <-- Why this ?????
2 + 3
5

我不明白如何评估 1 + 2 以计算结果 5。从理论上讲，我确实了解def应该强制重新计算fib，但我无法找到执行跟踪中可能发生这种情况的位置。

我想通过我的理解来引导你们

Output( My understanding):
1  
This is the head, trivial
2  
This is the tail of the first Cons in Cons( 1, Cons( 2, fn ) ). Trivial.
1 + 2
(fib zip fib.tail) map {case (x, y) => println("%s + %s".format(x, y)); x + y}))
first element of fib is 1
first element of fib.tail is 2
Hence 1 + 2 is printed.
The zip operation on the Stream does the following
 Cons( ( this.head, that.head), this.tail zip that.tail ) # this is fib and that is fib.tail. Also remember that this.tail starts from 2 and that.tail would start from 3. This new Stream forms an input to the map operation.
The map operation does the following 
cons(f(head), tail map f ) # In this case tail is a stream defined in the previous step and it's not evaluated.
So, in the next iteration when tail map f is evaluated shouldn't just 2 + 3 be printed ? I don't understand why 1 + 2 is first printed 

:( :( :(

我错过了什么明显的东西吗？

Answer 1

https://stackoverflow.com/a/20737241/3189923 年提出的斐波那契编码，此处添加了详细程度以跟踪执行，

val fibs: Stream[Int] = 0 #:: fibs.scanLeft(1)((a,b) => {
  println(s"$a + $b = ${a+b}")
  a+b
})

然后，例如，

scala> fibs(7)
1 + 0 = 1
1 + 1 = 2
2 + 1 = 3
3 + 2 = 5
5 + 3 = 8
8 + 5 = 13
res38: Int = 13