Ternary nesting -- bad idea?

This functions returns the n-th number in the fibonacci sequence. Besides crappy readability, is there any reason for not doing it this way?

Code:

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long unsigned int fibby(int n, long unsigned int num1 = 0LU, long unsigned int num2 = 1LU, int count = 1);
{
  return (!n || n == 1) ? 0 : ((count < n) ? fibby(n, num2, num2 + num1, count + 1) : num2);
}

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I've learned that the ternary can be very optimal, but does using it this way backfire that optimization? I don't think so, but tell we what you know. Just a thought.

whoops -- it's (n - 1):

Code:

---

long unsigned int fibby(int n, long unsigned int num1 = 0LU, long unsigned int num2 = 1LU, int count = 1);
{
  return (!n || n == 1) ? 0 : ((count < n - 1) ? fibby(n, num2, num2 + num1, count + 1) : num2);
}

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looks like crap

Quote:

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looks like crap

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lol

Try this, although this is more of a hack.

Code:

---

/*

  1/sqrt(5)  *  [1/2(1+sqrt(5)] ^n
 
  Approximates the nth term of the fibby sequence
  Works well when n >= 5
*/

#include<iostream>
#include<math.h>
#include<iomanip>

int fibby(int);

using namespace std;

int main()
{
    int enter;
    cout<<"Enter the nth term for the fibby sequence.";
    cout<<"Works well when n>=5:";
    cin>>enter;
   
    //call the function fibby pass in the variable 'enter'
    fibby(enter);
   
    int stop;
    cin>>stop;
}   



//function declaration
int fibby(int n)
{
    float partone;
    float parttwo;
    float partthree;
    float answer;
   
    partone=1/sqrt(5.0);
   
    parttwo=0.5*(1+sqrt(5.0));
   
    partthree=pow(parttwo,n);
    answer=(partone*partthree);
    int var;
    if(answer<10)
    {
        var=1;
    }
    if((answer>=10)&&(answer<=100))
    {
        var=2;
    }
    if  ((answer>=100)&&(answer<=1000))
    {   
        var=3;
    }
      if  ((answer>=1000)&&(answer<=10000))
    {   
        var=4;
    }
    if  ((answer>=10000)&&(answer<=100000))
    {   
        var=5;
    }
   
    cout<< setprecision(var)<<answer;
}

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You can condense the set precision bit but I couldn't be bothered.

:confused:

how about this:

Code:

---

unsigned long fib(int n, unsigned long current = 1, unsigned long prev = 0)
{
    return(n <= 1 ? current : fib(n - 1, prev + current, current));
}

int main()
{
    for(int i = 1; i <= 20; i++)
    {
        std::cout << fib(i) << std::endl;
    }
    return(0);
}

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I realise it doesn't answer your question, but I had fun ;)