Can a function call itself?

Hi all, I have to program a recursion to find the answer to a maths question, the recursion formula I am working with is:

f(a,b) = f(a,b-1) - f(a/g(b),b-1)

where g is an arbitary function that is easy to use

I also have the condition that f(a,0)=a

What is the best way to go about programming this kind of thing in C?

My initial idea was to have the function call itself if b wasn't 0, but that doesn't seem to work, will this method work or is there a much better SIMPLE way?

Code:

---


MainCL{
printf("%d", f(10,5));
}

int f(int a, int b){
if(b=0){
return a;}
else{
return f(a,(b-1))-f(a/2,(b-1));}
}

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Quote:

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Originally posted by ajm218

Code:

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MainCL{
printf("%d", f(10,5));
}

---

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What exactly is this?

Quote:

---

Originally posted by ajm218

Code:

---


int f(int a, int b){
if(b=0){
return a;}
else{
return f(a,(b-1))-f(a/2,(b-1));}
}

---

---

Your if statement is wrong. '=' is assigmnent, '==' is an equality test.

Quzah.

Sorry just to give a quick example of some input, so i gave f(10,5)

Code:

---


MainCL{
printf("%d", f(10,5));
}

int f(int a, int b){
if(b==0){
return a;}
else{
return f(a,(b-1))-f(a/2,(b-1));}
}

---

what is MainCL , i never heard of it.
is it just like int main()?
can someone please explaine?

Wish instructors did not do braindead things like #define MainCL ...

As far as recursion, it is easy to do in C, you need 2 things:
1. your algorithm
2. stopping condition to unwind your recursion

Also you don't want to recurse too deep, every call to itself takes up stack space in a form of return address and parameters, after so many calls you may run out of stack space, but that's just some technical detail.

Code:

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#include "stdio.h"

double factorial(double a)
{
/* Stopping condition */
  if (a < 1.0)
    return 1.0;

/* algorithm based on itself */
  return a * factorial(a-1.0);
}

int main(int argc, char **argv)
{
  printf("factorial(5)=%f", factorial(5.0));

  return 0;
}

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