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

我定义了一个函数来返回斐波那契流，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}))
}

功能正常，但看起来效率低下(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

所以，看起来函数从一开始就计算第 n 个斐波那契数.这是正确的吗?你会怎么修?

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，您每次都在调用该方法，并且每次该方法调用都会构造一个新的流.def 和 val 之间的区别 之前已经在 SO 上介绍过，所以这里就不详述了.如果您有 Java 背景，那应该很清楚.

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Scala 的另一个好处；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.

这篇关于如何在 Scala 中修复我的斐波那契流的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！