I’m trying to create a Fibonacci sequence using Scala’s Stream feature. Here’s what I came up with:
lazy val fibSeq: Stream[Int] = 0 #:: 1 #:: (fibSeq, fibSeq.drop(1)).zipped.map(_ + _)
But when I try to print the first 10 numbers:
fibSeq.take(10).foreach(println)
I get a StackOverflowError after printing 0 and 1. It looks like the zipped method isn’t playing nice with Stream.
Does anyone know why this is happening? Is there a way to make this work with infinite lists? Or should I use a different approach altogether for generating Fibonacci numbers in Scala?
I’m new to functional programming and Scala, so any explanations or alternative solutions would be really helpful. Thanks!
The issue you’re encountering is due to the eager evaluation of the zipped method. Streams are meant to be lazy, but zipped forces evaluation of both streams, causing infinite recursion. A more effective approach would be to use the zip method instead, which maintains laziness:
lazy val fibSeq: Stream[Int] = 0 #:: fibSeq.scanLeft(1)(_ + _)
This implementation uses scanLeft to generate the sequence, which is both concise and efficient. It starts with 0, then adds each element to the previous one to generate the next Fibonacci number. This should work without any stack overflow issues, allowing you to generate and print as many Fibonacci numbers as you need.
If you’re interested in diving deeper into functional programming patterns in Scala, I’d recommend exploring more about lazy evaluation and infinite sequences. It’s a powerful concept that can lead to elegant solutions for many mathematical and computational problems.
I’ve been working with Scala for a while now, and I’ve found that when dealing with infinite sequences like the Fibonacci series, it’s often helpful to use the unfold function from the Stream object. Here’s an approach that’s worked well for me:
lazy val fibSeq: Stream[BigInt] = Stream.unfold((BigInt(0), BigInt(1))) { case (a, b) =>
Some((a, (b, a + b)))
}
This method is both efficient and easy to understand. It starts with a tuple of (0, 1) and generates each subsequent number by adding the previous two. The unfold function takes care of the recursion for us, avoiding stack overflow issues.
I’ve used BigInt here to handle larger numbers, which is crucial for real-world applications where you might need to generate many Fibonacci numbers. This approach has served me well in several projects, and it’s quite performant even for large sequences.
hey, i had similar issues. try this:
lazy val fibSeq: Stream[BigInt] = BigInt(0) #:: BigInt(1) #:: fibSeq.zip(fibSeq.tail).map { case (a, b) => a + b }
works for me without stackoverflow. using BigInt coz it can handle larger nums. hope it helps!