need some help thanks in advance

I need to write an iterative function of the program I wrote as recursive.

I am a total rookie at this c programing game. I got my program working with a recursive function and it does what it is suppose to do.

Example: 6 and 15 = 3 ; 15 and 22 =1

I gave it a shot writing the program without the recursive function but it doesnt compute properly.

Can someone help me out with this.

Code:

---

/* HERE IS MY RECURSIVE PROGRAM IT WORKS PERFECTLY /*
/* #include <stdio.h>  */
/* int gcd (int a, int b);  */
/* int main()  */
/* {  */

/*  int a=0;  */
/*  int b=0;  */
/*  printf(" This program will compute greatest common divisor of two integers. \n");  */
/*  printf("Please enter two integers: \n\n"); */
/*  scanf("%d", &a); */
/*  scanf("%d", &b); */
/*  printf("\n The greatest common denominator of %d and %d is %d \n", a, b, gcd(a,b)); */
/*  return 0; */
/* }  */
    /* Recursive Function */
               
/*        int gcd (int a,int b)*/
/*        {  */
/*                int r;  */
/*                if ((r=a%b) == 0) */
/*                        return b; */
/*                else  */
/*                        return gcd(b, r); */
/*        } */

/*  HERE IS MY ATTEMPT TO DO THIS ITERATIVE FUNCTION */
#include <stdio.h>
int gcd (int a, int b);
int main()
{

  int a=0; 
  int b=0;
  int r=0;
  printf(" This program will compute greatest common divisor of two integers. \n");
  printf("Please enter two integers: \n\n");
  scanf("%d", &a);
  scanf("%d", &b);
  r=(a%b);
  printf("\n The greatest common denominator of %d and %d is %d \n", a, b, r);
  return 0;
}

---

Let's take a look at the recursive version (slightly modified).

Code:

---

int gcd_r(int a, int b)
{
  int r = a % b;
  if ( r )
  {
      return gcd_r(b, r);
  }
  return b;
}

---

As long as a % b is nonzero, it will recursively call itself -- or loop. The (recursive) loop takes b as a and r as b when it repeats. Or...

Code:

---

int gcd_i(int a, int b)
{
  int r = a % b; /* same */
  while ( r ) /* mostly the same */
  {
      a = b; /* replaces parameter 1 of recursive call */
      b = r; /* replaces parameter 2 of recursive call */
      r = a % b; /* replaces recursive call */
  }
  return b;
}

---

I understand how the recursive function works... I need to have help with the iteritive version of the program... making it non recursive.

for some reason it is mis calculating.

Those snippets were from actual working code.

Code:

---

int main()
{
  int a = 6, b = 15; 
  printf("gcd_r(%d, %d) = %d\n", a, b, gcd_r(a,b));
  printf("gcd_i(%d, %d) = %d\n", a, b, gcd_i(a,b));
  a = 15,b = 22; 
  printf("gcd_r(%d, %d) = %d\n", a, b, gcd_r(a,b));
  printf("gcd_i(%d, %d) = %d\n", a, b, gcd_i(a,b));
  return 0;
}

/* my output
gcd_r(6, 15) = 3
gcd_i(6, 15) = 3
gcd_r(15, 22) = 1
gcd_i(15, 22) = 1
*/

---

Perhaps you could try a closer look at what you are brushing aside. And think about what I wrote in regard to changing a recursive loop into an interative loop.

not brushing aside any of the help you offer. I am glad you responded.. i have been trying to get the code you posted to work. I think i understand. my problem now is I cant get it to print out the reminder. Trying this latest post of yours now.

thanks for the help

Okay i have tried and tried to put your code into my program with no luck getting a printout.

I need it to printout like this:

if i enter 15 22

it shows 15 22 and says the remainder is 1

Just like my recursive program at the top.

I am thinking if i am having problems with these easy ones, I am in big trouble laters on. This is just an excercise in the book. Trying to figure it out.

Well, to convert a recursive function to an iterative one, you need at least a loop of some sort. (Maybe two for loops would work.)

I just wanted to say thanks dave for the help with this program. It was driving me crazy. I worked most of yesterday on it and quit. This morning got up read your posts and for some reason I got it. The program is working perfectly. I even understand how and why it is working.

Thanks again for your help.