fib.cr

require "benchmark"
require "big"

def fib1(n)
  a = 0
  b = 1
  n.times { a, b = b, a + b }
  a
end

def fib2(n)
  a = 0
  b = 1
  n.times { a, b = b, a + b; nil }
  a
end

def fib3(n)
  a = 0
  b = 1
  while n > 0
    a, b = b, a + b
    n -= 1
  end
  a
end

# let's force it to not be optimized by LLVM. Thanks to @petr-fischer!
val = (ARGV[0]?.try &.to_i?) || 45

Benchmark.bm do |x|
  x.report { 1_000_000.times { fib1(val) } }
  x.report { 1_000_000.times { fib2(val) } }
  x.report { 1_000_000.times { fib3(val) } }
end

fib.py

import timeit

def fib(n):
  a = 0
  b = 1
  while n > 0:
    a, b = b, a + b
    n -= 1
  a


def million_fibs():
  for _ in range(1000000):
    fib(45)

print(timeit.timeit(lambda: million_fibs(), number=1))

fib.rb

require "benchmark"

def fib1(n)
  a = 0
  b = 1
  n.times { a, b = b, a + b }
  a
end

def fib2(n)
  a = 0
  b = 1
  n.times { a, b = b, a + b; nil }
  a
end

def fib3(n)
  a = 0
  b = 1
  while n > 0
    a, b = b, a + b
    n -= 1
  end
  a
end

Benchmark.bm do |x|
  x.report { 1_000_000.times { fib1(45) } }
  x.report { 1_000_000.times { fib2(45) } }
  x.report { 1_000_000.times { fib3(45) } }
end

Ehm, you're calling fib on a constant number. Isn't Crystal capable of doing compile-time evaluation on constants, meaning it doesn't do any run-time calculations?

I tried to modify it and parse fib argument from ARGV command line like this:

# ...
val = ARGV[0].to_i
puts val

Benchmark.bm do |x|
  x.report { a = 0_u128; 1_000_000.times { a += fib1(val) }; puts a }
  x.report { a = 0_u128; 1_000_000.times { a += fib2(val) }; puts a }
  x.report { a = 0_u128; 1_000_000.times { a += fib3(val) }; puts a }
end

and now it's at least 10x slower.

Thanks! I edited the post (and the code) accordingly.

tl;dr: points stand still.