Keep running program and 'int' problem

This is a discussion on Keep running program and 'int' problem within the C++ Programming forums, part of the General Programming Boards category; I'm using this program in C++ to calculate wind chill: Code: #include <cstdlib> #include <iostream> #include <math.h> using namespace std; ...

  1. #1
    Registered User
    Join Date
    Feb 2008
    Posts
    11

    Thumbs up Keep running program and 'int' problem

    I'm using this program in C++ to calculate wind chill:

    Code:
    #include <cstdlib>
    #include <iostream>
    #include <math.h>
    
    using namespace std;
    
    int main(int argc, char *argv[])
    {
        cout << "Wind Chill Calculator...\n" ;
        cout << "By William Seed\n" ;
        int temp, vel;
        int ch;
        cout << "Enter the temperature:";
        cin >> temp;
        cout << "Enter the wind speed:";
        cin >> vel;
        ch = 35.75 + (.6215 * temp) - (35.75 * pow(vel,.16)) + (.4275 * temp * pow(vel,.16));
        cout << "The wind chill is: ";
        cout << ch;
        cout << "\n";
        system("PAUSE");
        return EXIT_SUCCESS;
    }
    I have two problems:

    1. (most important) I want this program to keep running until I tell it to stop (using imput as yes or no.
    2. If I try to use a decimal for 'temp', then the program ends...

    I'm new, so no flaming, please.

    Thanks in advance,

    Will

  2. #2
    Registered User
    Join Date
    Feb 2008
    Posts
    11
    Never mind about the second problem. I fixed it by declaring it a 'long double'.

  3. #3
    Registered User NeonBlack's Avatar
    Join Date
    Nov 2007
    Posts
    435
    Your temperature, windspeed and chill variables should all be double or float.
    Since integers are whole numbers, using decimal points is inappropriate for the type int.

    To keep the program running, you need to use some kind of loop. When you finish everything, ask the user if she wants to run the program again, if she enters yes, go back to the top of the loop, if she says no, terminate the loop.

    It's up to you to figure out how to implement this.
    I copied it from the last program in which I passed a parameter, which would have been pre-1989 I guess. - esbo

  4. #4
    Its hard... But im here swgh's Avatar
    Join Date
    Apr 2005
    Location
    England
    Posts
    1,479
    Or use a sentinal controlled loop:

    Code:
    std::cout << "Enter a number or enter -1 to end: ";
    std::cin >> number;
    
    while ( number != -1 )
    {
       std::cout << "\nSqaured: " << number * number << std::endl;
    
       std::cout << "Enter a number or -1 to end: ";
       std::cin >> number;
    }
    This snippet will continue to square the number untill the user enters -1 to exit the loop
    I'm just trying to be a better person - My Name Is Earl

  5. #5
    Registered User NeonBlack's Avatar
    Join Date
    Nov 2007
    Posts
    435
    swgh I'm pretty sure this is his homework.
    I copied it from the last program in which I passed a parameter, which would have been pre-1989 I guess. - esbo

  6. #6
    Its hard... But im here swgh's Avatar
    Join Date
    Apr 2005
    Location
    England
    Posts
    1,479
    Oh sorry I did not complete the program for him I just mentioned a way to use a user controlled loop thats all.
    I'm just trying to be a better person - My Name Is Earl

  7. #7
    Registered User NeonBlack's Avatar
    Join Date
    Nov 2007
    Posts
    435
    Anyway, although the type of control swgh suggested can be useful, in my opinion there is a better loop for this case.
    Also, you should be careful when using != and == with floating point numbers.
    I copied it from the last program in which I passed a parameter, which would have been pre-1989 I guess. - esbo

  8. #8
    coder
    Join Date
    Feb 2008
    Posts
    127
    hi willrs2
    here is what you need for your loop implementation:
    http://www.cprogramming.com/tutorial/lesson3.html


    1st: I'd suggest you to divide your code in contextual blocks, in order to make it more readable, for us and for you too.

    2nd: if you need to print a line like
    Code:
    The wind chill is: <value>
    You may put all these data into a single row of code:
    Code:
    cout << "The wind chill is: " << ch << "\n";
    3rd: <iostream> provides a useful "endl" (end line) to let you avoid typing "\n" everytime:
    Code:
    cout << variable << endl;


    just an example:
    note the separated blocks of lines and where variable declarations should be put
    Code:
    int main(int argc, char *argv[])
    {
        int temp, vel;
        int ch;
    
        cout << "Wind Chill Calculator...\n" ;
        cout << "By William Seed\n" ;
    
        cout << "Enter the temperature:";
        cin >> temp;
    
        cout << "Enter the wind speed:";
        cin >> vel;
    
        ch = 35.75 + (.6215 * temp) - (35.75 * pow(vel,.16)) + (.4275 * temp * pow(vel,.16));
        cout << "The wind chill is: " cout << ch << endl;
    
        system("PAUSE");
        return EXIT_SUCCESS;
    }

  9. #9
    Registered User
    Join Date
    Feb 2008
    Posts
    11
    Thanks, all of you. I used a while loop. Thanks carlorfeo for the tips, and thanks swgh and NeonBlack for the loop idea... I completely forgot about it...

    Here is the code a the proggy working how I want it:

    Code:
    #include <cstdlib>
    #include <iostream>
    #include <math.h>
    
    using namespace std;
    
    int main(int argc, char *argv[])
    {
        double temp, vel;
        double ch;
        char yn;
        while (yn = 1) 
        {
              cout << "Wind Chill Calculator...\n" ;
              cout << "By William Seed\n" ;
              cout << "Enter the temperature: ";
              cin >> temp;
              cout << "Enter the wind speed: ";
              cin >> vel;
              ch = 35.75 + (.6215 * temp) - (35.75 * pow(vel,.16)) + (.4275 * temp * pow(vel,.16));
              cout << "The wind chill is: ";
              cout << ch;
              cout << "\n";
              cout << "Enter 1 to rerun program: " ;
              cin >> yn ;
              cout << "\n\n";
        }
    
        system("PAUSE");
        return EXIT_SUCCESS;
    }

    I have one more question, though... It's not that important, but if I enter anything besides 1, it still reruns the program. How could I fix that?
    Last edited by willrs2; 02-07-2008 at 05:23 PM.

  10. #10
    coder
    Join Date
    Feb 2008
    Posts
    127
    Code:
    while (yn = 1)
    this is an assignment, not a comparation

    comparation is done using ==
    Code:
    while (yn == 1)

  11. #11
    coder
    Join Date
    Feb 2008
    Posts
    127
    anyway your loop won't work since you tell the program to loop while yn is 1.
    Is it correct?

    edit:
    sorry, I didn't read the code well enough
    the real matter is that the loop won't start because, when you run the program, yn is not 1: its value has not been declared.

    Using while (yn = 1) worked beacause the statement (yn = 1), which is an assignment, will always result true.
    Last edited by carlorfeo; 02-07-2008 at 05:35 PM.

  12. #12
    Registered User
    Join Date
    Feb 2008
    Posts
    11
    It runs just fine the way it is in my previous post, but it won't exit.

    I only made two changes:

    I added in a second = to make it a comparison.

    and

    I set yn = 1 just before the while statement and after the char yn; statement, and it doesn't allow me to rerun the program at all.
    Last edited by willrs2; 02-07-2008 at 05:42 PM.

  13. #13
    Registered User
    Join Date
    Feb 2008
    Posts
    11
    I gotta go for now, maybe the day. Thanks for the help so far.

  14. #14
    Registered User
    Join Date
    Dec 2006
    Location
    Canada
    Posts
    3,183
    note the separated blocks of lines and where variable declarations should be put
    IMHO that is more of a personal preference. It is required in C but not C++. Some people recommend declaring variables just before they are used in C++.

  15. #15
    coder
    Join Date
    Feb 2008
    Posts
    127
    you are right cyberfish, I often declare variables not at beginning too.

    By the way, in that case these declarations stay well out of the loop
    Last edited by carlorfeo; 02-07-2008 at 07:21 PM.

Page 1 of 2 12 LastLast
Popular pages Recent additions subscribe to a feed

Similar Threads

  1. Debug Error Really Quick Question
    By GCNDoug in forum C Programming
    Replies: 1
    Last Post: 04-23-2007, 01:05 PM
  2. Personal Program that is making me go wtf?
    By Submeg in forum C Programming
    Replies: 20
    Last Post: 06-27-2006, 01:13 AM
  3. Contest Results - May 27, 2002
    By ygfperson in forum A Brief History of Cprogramming.com
    Replies: 18
    Last Post: 06-18-2002, 02:27 PM
  4. My Allegro Program will not run...
    By Vicious in forum Game Programming
    Replies: 17
    Last Post: 06-14-2002, 01:49 AM
  5. debug program
    By new_c in forum C Programming
    Replies: 3
    Last Post: 03-18-2002, 11:50 PM

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21