如何在Scala中修复我的斐波那契流 [英] How to fix my Fibonacci stream in Scala

问题描述

我定义了一个返回斐波那契流的函数，如下所示:

I defined a function to return Fibonacci stream as follows:

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}))
}

函数可以正常运行，但效率低下(请参见下面的输出)

The functions work ok but it looks inefficient (see the output below)

scala> fib take 5 foreach println
1
2
1 + 2
3
1 + 2
2 + 3
5
1 + 2
1 + 2
2 + 3
3 + 5
8

因此，函数似乎从一开始就计算第n个斐波那契数.这是正确的吗?您将如何解决?

So, it looks like the function calculates the n-th fibonacci number from the very beginning. Is it correct? How would you fix it?

推荐答案

那是因为您使用了 def .尝试使用 val :

That is because you have used a def. Try using a val:

lazy val fib: Stream[Int] 
  = 1 #:: 2 #:: (fib zip fib.tail map { case (x, y) => x + y })

基本上， def 是一种方法；在您的示例中，每次调用该方法时，每次方法调用构造一个新流时，都将调用该方法. def 和 val 之间的区别在之前已在SO上进行了覆盖，因此在此不再赘述.如果您来自Java背景，那么应该很清楚.

Basically a def is a method; in your example you are calling the method each time and each time the method call constructs a new stream. The distinction between def and val has been covered on SO before, so I won't go into detail here. If you are from a Java background, it should be pretty clear.

这是关于scala的另一件事.在Java中，方法可能是递归的，但类型和值可能不是.在Scala中，值和类型都可以递归.

This is another nice thing about scala; in Java, methods may be recursive but types and values may not be. In scala both values and types can be recursive.

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