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Which is the best approach to generate fibonacci

Two common approaches for generating the Fibonacci series:

  • traditional approach, i.e. starting a for loop inside a function.
  • Recursion

I met another solution

#include <iostream>

using namespace std;

void fibo() {
 static int y = 0;
 static int x = 1;
 cout << y << endl;
 y = x + y;
 x = y - x;
}

int main() {
 for (int i = 1; i <= 1; i++) {
  fibo();
 }
 return 0;
}

This solution works fine in initial launches, but compared to the traditional and recursive approach, does it have any significant flaws?

I am sure that static variables would add to the complexity of space, but at least we are not building a stack of function tables using recursion, right?

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6 answers

The disadvantages that I can immediately see:

  • Essentially making the state global, it is not thread safe
  • , reset

, , - , . ( , ++, # Java IEnumerable<T> Iterable<T>.)

+7

, , , (, , - , ), , , . , , , . :

class FiboNumbers {
  public:
    FiboNumbers() :
        x_(1), y_() {}

    int getNext() {
        x_ += y_;
        y_ = x_ - y_;
        return x_;
    }

  private:
    int x_, y_;
};

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+2

++ ( 1,5 ).

, .

#include<iostream>

using namespace std;

void fibseries(long int n)
{
    double x=0;double y=1;
    for (long int i=1;i<=n;i++)
     {
        if(i%2==1)
         {
            cout<<x<<" ";
            x=x+y;
         } 
        else 
         {
            cout<<y<<" ";
            y=x+y;
         }
     }
}

main()
{
    long int n=0;
    cout<<"The number of terms ";
    cin>>n;
    fibseries(n);
    return 0;
}
+1

, . , , , , . ( , , , "0" .) :

int y = 0;
int x = 1;
for ( int i = 0; i < count; ++ i ) {
    std::cout << y <<std::endl;
    y = x + y;
    x = y - x;
}

? , .

0

, , n- , n-1- .

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class, Sevis, , : , ( ), , , ..

0
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I think this pointer approach will be more useful for you.

void main()
{
    int i,p, *no,factorial,summ;

    int fib(int p);

    clrscr();

    printf("\n Enter The Number:");
    scanf("%d", no);
    printf("\n The Fibonnacci series: \n");

    for(i=0; i < *no; i++)
        printf("%d\n", fib(i));

    getch();
}

int fib(int p)
{
    if(p == 0)
        return(0);
    if(p >= 1 && p <= 2)
        return(1);
    else
        return(fib(p - 1) + fib(p - 2));
}
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