Go Premium for a chance to win a PS4. Enter to Win

x
?
Solved

Using a do-while loop and function  calls

Posted on 2000-03-14
2
Medium Priority
?
1,798 Views
Last Modified: 2010-08-05
main()
{
    int a = 0, b = 1, c = 0               /* initialize the series to 0, 1 */
    int sum();                              /* declare function to calc sums */

    printf("fibonacci\n");
    printf("%5d \n", a);
    printf("%5d \n", b);

    do
    {

     c = sum(a, b);
     printf("%5d\n", c);
     a = b;
     b = c;

     }
     
     while (c < 10000);

}

int sum (x, y) /* the parameters x and y receive the values of */
int x, y;          /* the two arguments passed by sum(a, b) in main() */

{

    return x + y;

}


Question: I need to use a subtract function to calculate and display the differences between the suceeding numbers in the fibonacci series.
I need to calculate and display thr quotients of the succeding fibonacci series in a third data column, right next to the differences.  For the divisions I need a division function named divide to perform this calculation.  To preserve the decimal accuracy of thequotients I will need to declare and use a float type function with two arguments  
0
Comment
Question by:mvjohn
2 Comments
 
LVL 7

Expert Comment

by:KangaRoo
ID: 2615611
When the new number is calculated, the old number remains in b, so the difference could be obtained from c - b, which is probably equal to a, provided that sum(a,b) calculates a + b ... The quotient is double(c) / b. I don't see why you would need functions for these simple expressions
0
 
LVL 10

Accepted Solution

by:
RONSLOW earned 30 total points
ID: 2619912
KangaRoo: Because it is a school/college/uni assignment? :-)

mvjohn: for accuracy, use a double, not a float type.  float types are not much use (except to save memeory when you are using a LOT of them).

Also, you are using yucky old-style C declarations.  This is bad programming practice.  You should declare your sum function as:

  int sum (int x, int y);

and define it as:

  int sum (int x, int y) {
    return x+y;
  }

I think, without giving too much away, your subtract and divide functions would be

  int subtract (int x, int y) {
    return x-y;
  }

  double divide (double x, double y) {
    return x/y;
  }

then you'd change
  printf("%5d\n", c);
into
  printf("%5d %5d %8.5f\n", c, subtract(c,b), divide(c,b));

PS: you'll probaby want to put some error checking in the divide routine to avoid division by zero.


0

Featured Post

Free Tool: Subnet Calculator

The subnet calculator helps you design networks by taking an IP address and network mask and returning information such as network, broadcast address, and host range.

One of a set of tools we're offering as a way of saying thank you for being a part of the community.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

When writing generic code, using template meta-programming techniques, it is sometimes useful to know if a type is convertible to another type. A good example of when this might be is if you are writing diagnostic instrumentation for code to generat…
  Included as part of the C++ Standard Template Library (STL) is a collection of generic containers. Each of these containers serves a different purpose and has different pros and cons. It is often difficult to decide which container to use and …
The goal of the video will be to teach the user the concept of local variables and scope. An example of a locally defined variable will be given as well as an explanation of what scope is in C++. The local variable and concept of scope will be relat…
The viewer will learn how to pass data into a function in C++. This is one step further in using functions. Instead of only printing text onto the console, the function will be able to perform calculations with argumentents given by the user.
Suggested Courses

876 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question