Following java program creates an infinite fibonacci sequence (stream) an takes the first 42 values of it.

import java.util.function.UnaryOperator;
import java.util.stream.IntStream;
import java.util.stream.Stream;

public class test3 {
public static void main(String[] args) {
int beginvalue[] = new int[] { 0, 1 };
UnaryOperator<int[]> fibo_seq = a -> new int[] { a[1], a[0] + a[1] };
Stream<int[]> s = Stream.iterate(beginvalue, fibo_seq);
IntStream fibStream = s.mapToInt(a -> a[1]);
IntStream limitedstream = fibStream.limit(42);
limitedstream.forEach(System.out::println);
}
}

How would this program look converted to dlang ?
Maybe there are multiple solutions ?

On Thursday, 21 April 2022 at 03:41:24 UTC, Ali Çehreli wrote:

My favorite is the struct range. Because it is more understandable and personalized. Moreover, you can limit it without using take().

struct FibonacciRange(int l)
{
  long a = 1, b = 1;

  bool empty()
  {
    return a > l;
  }

  long front() const
  {
   return a;
  }

  void popFront()
  {
    auto t = a;
    a = b;
    b += t;
  }
}

enum limit = 16;

void main()
{
  import std.stdio;
  FibonacciRange!limit fibs;
  fibs.write(": ");

  import std.algorithm;
  auto total = fibs.sum;
  total.writeln;

  assert(total == 33);
}

SDB@79

On Thursday, 21 April 2022 at 04:36:13 UTC, Salih Dincer wrote:

struct FibonacciRange(int l)
{
  long a = 1, b = 1;

  bool empty()
  {
    return a > l;
  }

  long front() const
  {
   return a;
  }

  void popFront()
  {
    auto t = a;
    a = b;
    b += t;
  }
}

enum limit = 16;

void main()
{
  import std.stdio;
  FibonacciRange!limit fibs;
  fibs.write(": ");

  import std.algorithm;
  auto total = fibs.sum;
  total.writeln;

  assert(total == 33);
}

This example limits the maximum value returned by the fibonacci function. f(n) < limit
But it does not allow to return the n-th element of a fibonacci function.

On Thursday, 21 April 2022 at 05:00:53 UTC, Alain De Vos wrote:

You are free to use take():

struct FibonacciRange(long l) { /*...*/ }

enum limit = long.max;

void main()
{
  FibonacciRange!limit fibs;
  auto fibFirst90 = fibs.take(90);
  auto total = fibFirst90.sum;
       total.writeln; // "7540113804746346428" == fib(92)-1

  auto fibInfinite = recurrence!("a[n-1] + a[n-2]")(1L, 1L);
  assert(fibInfinite.take(92).array[$-1]-1 == total);
}

SDB@79

On Thursday, 21 April 2022 at 04:36:13 UTC, Salih Dincer wrote:

And it's inherently lazy, so no extra processing/calculation other than what's requested.