Fibonacci with recursion

[Tired]

hey..
can u help me wid this..
#include<stdio.h>
#include<conio.h>
void main()
{
int fib(int);
int n,f=1,ans;
clrscr();
while(f==1)
{
printf("\n***Generating A Fibonacci Series***");
printf("\nEnter Number Of Terms Required: ");
scanf("%d",&n);
if(n<=1)
{
printf("\nInvalid Choice..Number Of Terms Should Be more Than 1");
printf("\nDo You Want To Try Again...\n1.Yes\n2.No");
scanf("%d",&f);
}
else
{
printf("\nSeries: %d",fib(n));
f=2;
}
}
getch();
}
int fib(int a)
{
if((a==0)||(a==1))
{
return(1);
}
else
{
return(fib(a-1)+fib(a-2));
}
}
i wat to print d numbers as well...eg if n =5
1 1 2 3 5

[newb16]

@Tired
You what to what?

void fibfib(int n)
{
int q;
if(q=n-1)fibfib(q);printf("%d ",fib(q));
}

[Tired]

wat=*want...
umm i dint understand the func u hv typed..can u help me with understanding it???it wud b great if u can..

thnx..

[JosAH]

@Tired
Please spell your words properly; this is an international forum not your local text speak community.

kind regards,

Jos

[Tired]

I am sorry...but can u please help me understand this...
void fibfib(int n)
{
int q;
if(q=n-1)fibfib(q);printf("%d ",fib(q));
}

[JosAH]

@Tired
Note that fib() is your function; the function fibfib() is just defined for printing the return values of your function. Try it by hand for small values of n. Also note that the expression q=n-1 is non-zero when n != 1

kind regards,

Jos

[Tired]

But its not printing all the values(or the entire series). Its just printing the final answer.Where should i call this function?in the main block before i am printing my answer?
And most of the times its not printing any thing.

[JosAH]

@Tired
How strange; to be sure I ran it as well and it runs fine, i.e. it prints out the first n values of the standard Fibonacci series. Can you show us some code?

kind regards,

Jos

[Tired]

#include<stdio.h>
#include<conio.h>
void main()
{
int fib(int);
void fibfib(int);
int n,f=1,ans;
clrscr();
while(f==1)
{
printf("\n***Generating A Fibonacci Series***");
printf("\nEnter Number Of Terms Required: ");
scanf("%d",&n);
if(n<=1)
{
printf("\nInvalid Choice..Number Of Terms Should Be more Than 1");
printf("\nDo You Want To Try Again...\n1.Yes\n2.No");
scanf("%d",&f);
}
else
{
fibfib(n);
printf("\nAns i.e. term no. %d = %d",n,fib(n));
printf("\nDo You Want To Continue:\n1.Yes\n2.No\n");
scanf("%d",&f);
}
}
}
int fib(int a)
{
if((a==0)||(a==1))
return(1);
else
return(fib(a-1)+fib(a-2));
}
void fibfib(int a)
{
int q;
if(q=a-1)
fibfib(q);
else
printf("%d ",fib(q));
}

this is the code i tried.....

[JosAH]

@Tired
I don't see an 'else' in the original code ...

kind regards,

Jos

[Tired]

it doesnt work without else also!!!i tried..so then is that last print statement not a part of the for loop??and is the way i am calling the function right??

[JosAH]

@Tired
Don't shout; there is no need to start a "yes it does, not it doesn't" discussion. You have to show or prove that it doesn't work; at least show the code you used. I also ran it and it worked for me (both in C and C++, my Java version also ran correctly).

kind regards,

Jos

[khaichiew85]

i have tried ur code..no problem with that.
But i would like to advise u that without detecting if the user want to continue by pressing 1 or Y or 2 or N.Use isdigit to differentiate them else if when alphabet is pressed the loop will keep on going

[Tired]

hey khaichiew85,
so u mean that the "fibfib" func is displaying the series for u?can u help with the calling statement and evrything??

regards
Amy

[JosAH]

@Tired
There is no need for help: I simply copied/pasted that fibfib function, crafted a loop around it and voila. Why don't you show your code? We are not going to spoonfeed you complete programs.

kind regards,

Jos (moderator)