Two versions of fibonacci, null expression

Question

Here are two versions of fibonacci I thought of

def fibonacci( n )
  return  n  if ( 0..1 ).include? n
  ( fibonacci( n - 1 ) + fibonacci( n - 2 ) )
end
puts fibonacci( 5 )


def fib(n)
  if n == (0..1)
    return n
  elsif 
    return fib(n-1) + fib(n-2)
  end
end
puts fib(5)

Why doesent the second fib(n) work?

sawa · Accepted Answer · 2016-06-27T15:12:45.610

2

It is because n == (0..1) is never satisfied if n is a number. The range 0..1 is not a number. A minimum fix is:

(0..1) === n

edited Jun 27 '16 at 15:12

answered Jun 27 '16 at 15:11

sawa

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I will delete my answer if it is offending as much as your comment is. – sawa Jun 27 '16 at 15:13
No, I am genuinely confused. No need to take offence – Muntasir Alam Jun 27 '16 at 15:14
@cresjoy this case `===` is like `include?`, check [here](http://stackoverflow.com/a/4467823/1301840) – lusketeer Jun 27 '16 at 15:18
I get a void value expression error on return fib(n-1) + fib(n-2), I will check out that link though thanks – Muntasir Alam Jun 27 '16 at 15:19
Now it tells me the stack level is too deep, I think im in unstoppable recursion – Muntasir Alam Jun 27 '16 at 15:31
any thoughts? @sawa – Muntasir Alam Jun 27 '16 at 15:39
His problem is that the direction of the `===` was wrong. – danielrsmith Jun 27 '16 at 15:41
@danielrsmith Not only that. The OP hasn't used any. – sawa Jun 27 '16 at 15:41
1

@sawa I get that, my hunch is that the OP simple switched the `==` with `===` rather than having it be the exact way your answer was posted. Since `===` is a method on the first operand you have to put the range first, or it doesn't know how to do the operation properly (using the numeric version). Upvoted since this is correct answer. – danielrsmith Jun 27 '16 at 15:45

score 2 · Answer 2 · answered Jun 27 '16 at 15:40

2

You need to remove the elsif in favor of else (since you have no case), and also reverse the ===

def fib(n)
  if (0..1) === n
    return n
  else
    return fib(n-1) + fib(n-2)
  end
end
puts fib(5)

answered Jun 27 '16 at 15:40

danielrsmith

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There is no "the" `===`. – sawa Jun 27 '16 at 15:41
Why does === need to be the other way around? Why does double equal not work? – Muntasir Alam Jun 27 '16 at 15:43
`a === b` is just syntactic sugar for `a.===(b)`. So the object `a` is having the `===` method run. A number's `===` method is an equality comparison. On the other hand `(0..1)` is a `Range` object, and it's `===` method is a "include" type operation: http://ruby-doc.org/core-2.2.0/Range.html#method-i-3D-3D-3D . `==` won't work because as the documentations states, it needs to be the exact same range. – danielrsmith Jun 27 '16 at 15:48
@cresjoy you should accept sawa's answer it was correct. – danielrsmith Jun 27 '16 at 15:52
May I ask why, when I did n == (0..1), it gave me a stack overflow error? I know this is syntatically wrong. But that error gives the impression of infinite recursion. – Muntasir Alam Jun 27 '16 at 15:59
1

Since `n == (0..1)` never evaluated to true, you had no condition to stop the recursion. – danielrsmith Jun 27 '16 at 16:00

Two versions of fibonacci, null expression

2 Answers2