Looping a code

hi, i want to loop my code, where if the choice is 1, it will print the results then i want to ask the user if he/she wants to run the code again. if answered yes, the code will run again from the start. thank you

Code:

---

void main()
{
 char  emp[30], id[10];
 int  days, ot;
 float rate, bp, oot, gp, np, sss, phil, pag;
 int  x, y, choice;

 clrscr();

 g(5,15);  p("Employee's name: ");
 g(5,16);  p("ID number: ");
 g(5,17);  p("Rate per day: ");
 g(50,15); p("Days: ");
 g(50,16); p("Overtime: ");

 g(22,15); gets(emp);
 g(22,16); gets(id);
 g(22,17); s("%f", &rate);
 g(60,15); s("%d", &days);
 g(60,16); s("%d", &ot);

  gp  = grosspay(rate, days, ot);
  sss = ssscom(gp);
  pag = pagcom(gp);
  phil= philcom(gp);
  np  = netpay(gp, sss, phil, pag);

 g(5,20);  p("Your net pay is: %.3f ", np);
 g(5,22);  p("Do you want to see the breakdown?  [1] YES  [2] NO : ");
 g(58,22); s("%d", &choice);


 if(choice==1)
 {

      bp = rate*days;
      oot= (1.25*(rate/8))*ot;


      g(28,27); p("EMPLOYEE'S WAGE INFORMATION");
      g(5,29);  p("Employee:      %s", emp);
      g(5,30);  p("ID number:    %s", id);
      g(5,31);  p("Rate per day:  P %.2f", rate);
      g(50,29); p("Days:    %d", days);
      g(50,30); p("Overtime: %d", ot);

      g(20,35); p("-EARNINGS-");
      g(5,37);  p("Basic pay: P %.3f", bp);
      g(5,38);  p("Overtime:  P %.3f", oot);
      g(5,41);  p("Grosspay:  P %.3f", gp);

      g(50,35); p("-DEDUCTIONS-");
      g(38,37); p("SSS:        P %.3f", sss);
      g(38,38); p("PAGIBIG:    P %.3f", pag);
      g(38,39); p("PhilHealth: P %.3f", phil);
      g(38,41); p("Netpay:    P %.3f", np);

      This is where i want to ask if a user wants to run this block of code again, if answered YES.

 }


 else if(choice==2)
 {

      clrscr();
      for(x=31;x<=51;x++)
      {
        g(x,23); p("Í");
        g(x,25); p("Í");
      }

      g(31,24); p("º");
      g(51,24); p("º");
      g(31,23); p("É");
      g(51,23); p("»");
      g(31,25); p("È");
      g(51,25); p("¼");
      g(32,24); p("Thank you, goodbye!");
 }

 else
 {
      clrscr();
      g(33,24); p("Invalid Input!");

  Here also
 }


 getch();
}

---

please help, i need this tomorrow. thank you

I don't know what language this is, but in C, you need to declare your variables. You have not declared g, p, or s, in your program.

It won't link.

If you need it by tomorrow, you should have posted it before 10:51 pm the night before.

this is c, i just not included the #includes and #defines, i just posted the main() because i just need to loop the code. p is printf, s is scanf and g is gotoxy, and it is only 6:47pm here in the Philippines =)

Add a do/while loop, a new variable to hold the user's answer, and a printf, scanf and compare of the user's answer to your question and you'll be good to go.

this is what i did:

Code:

---

main()
{

clrscr();

do {

if(conditions)
{
  some prints

 printf("Again?");
 scanf("%d", &z);
}

else if(condition)
{
 some prints
}

else
{
  some prints

 ANOTHER ASKING IF WANT TO RE-RUN THE WHOLE CODE
}

} while(z==1);

getch();
}

---

am i doing this correctly? but the problem is, when i ended up in else and asks again to run the code, i cannot input anything and the program ends. also another thing, when it loops, i cannot get back to input on Employee's name, it skips to 2nd line where i need to input the ID. thanks you

Quote:

---

hi, i want to loop my code, where if the choice is 1, it will print the results then i want to ask the user if he/she wants to run the code again. if answered yes, the code will run again from the start. thank you

---

The key here is a structured approach.

Step 1. Dissecting your requirements:

* Apparently your code does something
* You want to print results when user enters a particular symbol
* You want to loop when user enters a particular symbol

Step 2. Translate to flow diagram:

Code:

---


+--------------+
| Do something |<-------------------------------+
+--------------+                                |
        |                                      |
        v                                      |
+----------------+                              |
| Get user input |                              |
+----------------+                              |
        |                                      |
        v                                      |
        *                                      |
      / \                                      |
      /  \                                    |
    /    \                                    |
    /      \                                  |
  /        \                                  |
  /          \  YES  +---------------+      |
 * input = 1?  * -----> | Print results |      |
  \          /        +---------------+      |
  \        /                  |              |
    \      /                    |              |
    \    /                    |              |
      \  /                      |              |
      \ /                      |              |
        *                        |              |
        |                        |              |
    NO +------------------------+              |
        |                                      |
        v                                      |
        *                                      |
      / \                                      |
      /  \                                    |
    /    \                                    |
    /      \                                  |
  /        \                                  |
  /          \  YES                          |
 * input = y?  * -------------------------------+
  \          /       
  \        /
    \      /
    \    /
      \  /
      \ /
        *
    NO |
        v
    +------+
    | Done |
    +------+

---

Notice how, up until this point, it does not matter whether you speak C, Python, Perl, or whatever. This is simply a description of how your program is supposed to work. Once you have this drawn up, the actual implementation shouldn't be too hard now.

Step 3. Translate to pseudo-code, or a skeleton framework:

Code:

---

while(!done)
{
    result = do_something();

    key = get_user_input();
    if(key == 1)
        print(results);

    key = get_user_input();
    if(key != y)
        done = true;
}

---

Step 4. Translate to C:

Up to you.

Stick to this general approach for all your projects, and you will succeed.

Additional note: USE PROPER INDENTATION. There is absolutely no excuse for not doing so.

wow thanks for that, ok i will. thanks for the hard work =)