C++ Recursion function explained using Fibonacci series

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A function can call itself. This is called recursion, and recursion can be direct or indirect. It is direct when a function calls itself; it is indirect recursion when a function calls another function that then calls the first function.

Some problems are most easily solved by recursion, usually those in which you act on data and then act in the same way on the result. Both types of recursion, direct and indirect, come in two varieties: those that eventually end and produce an answer, and those that never end and produce a runtime failure. Programmers think that the latter is quite funny (when it happens to someone else).

It is important to note that when a function calls itself, a new copy of that function is run. The local variables in the second version are independent of the local variables in the first, and they cannot affect one another directly, any more than the local variables in main() can affect the local variables in any function it calls, as was illustrated in Listing 5.4.

To illustrate solving a problem using recursion, consider the Fibonacci series:


Each number, after the second, is the sum of the two numbers before it. A Fibonacci problem might be to determine what the 12th number in the series is.

One way to solve this problem is to examine the series carefully. The first two numbers are 1. Each subsequent number is the sum of the previous two numbers. Thus, the seventh number is the sum of the sixth and fifth numbers. More generally, the nth number is the sum of n - 2 and n - 1, as long as n > 2.

Recursive functions need a stop condition. Something must happen to cause the program to stop recursing, or it will never end. In the Fibonacci series, n < 3 is a stop condition.

The algorithm to use is this:

1. Ask the user for a position in the series.

Call the fib() function with that position, passing in the value the user entered.

The fib() function examines the argument (n). If n < 3 it returns 1; otherwise, fib() calls itself (recursively) passing in n-2, calls itself again passing in n-1, and returns the sum.

If you call fib(1), it returns 1. If you call fib(2), it returns 1. If you call fib(3), it returns the sum of calling fib(2) and fib(1). Because fib(2) returns 1 and fib(1) returns 1, fib(3) will return 2.

If you call fib(4), it returns the sum of calling fib(3) and fib(2). We've established that fib(3) returns 2 (by calling fib(2) and fib(1)) and that fib(2) returns 1, so fib(4) will sum these numbers and return 3, which is the fourth number in the series.

Taking this one more step, if you call fib(5), it will return the sum of fib(4) and fib(3). We've established that fib(4) returns 3 and fib(3) returns 2, so the sum returned will be 5.

This method is not the most efficient way to solve this problem (in fib(20) the fib() function is called 13,529 times!), but it does work. Be careful: if you feed in too large a number, you'll run out of memory. Every time fib() is called, memory is set aside. When it returns, memory is freed. With recursion, memory continues to be set aside before it is freed, and this system can eat memory very quickly. Sample program below implements the fib() function.


WARNING: When you run this program, use a small number (less than 15). Because this uses recursion, it can consume a lot of memory.

Demonstrates recursion using the Fibonacci series.

1:     // Demonstrates recursion
2:     // Fibonacci find.
3:     // Finds the nth Fibonacci number
4:     // Uses this algorithm: Fib(n) = fib(n-1) + fib(n-2)
5:     // Stop conditions: n = 2 || n = 1
7:     #include <iostream.h>
9:     int fib(int n);
11:    int main()
12:    {
14:      int n, answer;
15:      cout << "Enter number to find: ";
16:      cin >> n;
18:      cout << "\n\n";
20:      answer = fib(n);
22:      cout << answer << " is the " << n << "th Fibonacci number\n";
23:         return 0;
24:    }
26:    int fib (int n)
27:    {
28:      cout << "Processing fib(" << n << ")... ";
30:      if (n < 3 )
31:      {
32:         cout << "Return 1!\n";
33:         return (1);
34:      }
35:      else
36:      {
37:        cout << "Call fib(" << n-2 << ") and fib(" << n-1 << ").\n";
38:        return( fib(n-2) + fib(n-1));
39:      } 
40: }

Output: Enter number to find:  5

Processing fib(5)... Call fib(3) and fib(4).
Processing fib(3)... Call fib(1) and fib(2).
Processing fib(1)... Return 1!
Processing fib(2)... Return 1!
Processing fib(4)... Call fib(2) and fib(3).
Processing fib(2)... Return 1!
Processing fib(3)... Call fib(1) and fib(2).
Processing fib(1)... Return 1!
Processing fib(2)... Return 1!
5 is the 5th Fibonacci number

Analysis: The program asks for a number to find on line 15 and assigns that number to target. It then calls fib() with the target. Execution branches to the fib() function, where, on line 28, it prints its argument.

The argument n is tested to see whether it equals 1 or 2 on line 30; if so, fib() returns. Otherwise, it returns the sums of the values returned by calling fib() on n-2 and n-1.

In the example, n is 5 so fib(5) is called from main(). Execution jumps to the fib() function, and n is tested for a value less than 3 on line 30. The test fails, so fib(5) returns the sum of the values returned by fib(3) and fib(4). That is, fib() is called on n-2 (5 - 2 = 3) and n-1 (5 - 1 = 4). fib(4) will return 3 and fib(3) will return 2, so the final answer will be 5.

Because fib(4) passes in an argument that is not less than 3, fib() will be called again, this time with 3 and 2. fib(3) will in turn call fib(2) and fib(1). Finally, the calls to fib(2) and fib(1) will both return 1, because these are the stop conditions.

The output traces these calls and the return values. Compile, link, and run this program, entering first 1, then 2, then 3, building up to 6, and watch the output carefully. Then, just for fun, try the number 20. If you don't run out of memory, it makes quite a show!

Recursion is not used often in C++ programming, but it can be a powerful and elegant tool for certain needs.

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1. View Comment

the program is writen beautifly and it is easy to understand.no difficult code is used......

View Tutorial          By: sania cheema at 2009-05-04 12:14:19
2. View Comment

i want simple program of function used in recursion,fibonacci.

View Tutorial          By: manish at 2009-05-19 01:03:19
3. View Comment

you example does not run it errors can you pls fixed it so that we can learn clearly

View Tutorial          By: dionesia at 2009-08-24 05:27:37
4. View Comment

it worked fine in solving an algorithm assingment.

View Tutorial          By: tunde bello at 2009-11-05 06:31:28
5. View Comment

Simple Codes please... I just wonder why does my program already had a value of 1 even I dont input any number...

View Tutorial          By: Joseph at 2009-11-12 05:37:50
6. View Comment

not satisfied...

want to print the fibonacci series

View Tutorial          By: nikhil jha at 2010-03-08 04:25:50
7. View Comment

for the guys who is facing problems
just change #include <iostream.h>
to #include<iostream>
and add
using namespace std;
it will work fine

View Tutorial          By: Elie at 2010-05-03 12:12:48
8. View Comment

it is very easy to understand to the learners.

View Tutorial          By: venkat at 2010-06-19 02:37:36
9. View Comment

You have clear my confusion by using simple codes!!!
Thank you

View Tutorial          By: salahu.kt at 2010-09-28 07:33:55
10. View Comment

how about this?...

int fibonacci(int n);

int main()

int number;

cout <<"Enter an integer :";
cin >> number;

for(int i=0;i<=number;i++)
cout <<fibonacci(i) <<" ";

cout << endl << endl;
cout <<"Fibonacci "<< i <<" is "<< fibonacci(i) <<endl;

return 0;

int fibonacci(int n) {
if (n <= 1) {
return n;
} else {
return fibonacci(n-1)+fibonacci(n-2);

View Tutorial          By: nadia at 2010-10-04 21:48:04
11. View Comment

my answer for this Question is :

#include <iostream>
using namespcae std;

in fib (int n)
if (n==0)
return 1;
return fib(n-1)+fib(n-2);

int main()
int x;
cout<<"please enter a number"<<endl;
cout<<"The fibonacci of "<<x<<"is"<<fib(x);

return 0;

View Tutorial          By: Tahani at 2011-04-26 11:55:34
12. View Comment

# include <iostream.h>
void main()
int i,ary[10];
ary[0] = 0;
ary[1] = 1;
ary[i] = ary[i-2] + ary [i-1] ;
cout<<" "<< ary[i];

View Tutorial          By: Ali Jafar at 2011-09-18 03:47:47
13. View Comment

give me some logics for programs as soon as possible

View Tutorial          By: nari at 2011-12-07 18:28:03

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