Statement that will tell the program to end?

[PYROMANIAC702]

I'm a noob at C and I need to know how to end the program in the middle so it won't do any more steps. I'm not using loops so break won't work. Please help!

[stahta01]

Look up exit() function; IRCC in the stdlib.
Note: make sure the return statement is not enough to solve your problem before using exit.

Edit: C++ docs on exit http://www.cplusplus.com/reference/c.../cstdlib/exit/

Tim S.

[PYROMANIAC702]

Look up exit() function; IRCC in the stdlib.
Note: make sure the return statement is not enough to solve your problem before using exit.

Tim S.

Thanks, but the exit statement isn't working. Here is my code:

#include <stdio.h>
#include <math.h>
#include "genlib.h"
#include "simpio.h"

main()

{
      int a, b, c, X, D, x1, x2;
      
      printf ("This program solves a quadratic equation.\n");
      printf ("Enter the value of A: ");
      a = GetInteger();
      printf ("Enter the value of B: ");
      b = GetInteger();
      printf ("Enter the value of C: ");
      c = GetInteger();
      if (a = 0)
      if (b = 0)
      {
            printf ("no solution.");
            exit();
            }
            else
            {
                X = - c / b;
                printf ("The equation is not quadratic and the solution is: %d\n", X);
                exit();
                }
      D = b * b - 4 * a * c;
      if (D < 0)
      {
            printf ("There is no real solution.\n");
            exit();
            }
            if (D = 0)
            {
                  x1 = - b / 2 * a;
                  printf ("One solution: %d", x1);
                  exit();
                  }
      x1 = (- b + sqrt(D)) / 2 * a;
      x2 = (- b - sqrt(D)) / 2 * a;
      printf ("Two solutions: %d and %d", x1, x2);
      getchar();
      }

[Matticus]

From the link provided by stahta01:

 void exit ( int status );

status
Status value returned to the parent process. Generally, a return value of 0 or EXIT_SUCCESS indicates success, and any other value or the constant EXIT_FAILURE is used to indicate an error or some kind of abnormal program termination.

Error from my compiler:

 insufficient number of arguments to 'exit'

Now put them together, and you'll find the answer.

[hk_mp5kpdw]

There is a difference between = (assignment) and == (equality test). All your "if" statements need to be using == and not =.

Also, you should #include the stdlib header if you want to use exit.

[laserlight]

You do not need exit here: you should either structure your code such that returning at the end of main will do, or use return statements.

[PYROMANIAC702]

There is a difference between = (assignment) and == (equality test). All your "if" statements need to be using == and not =.

Also, you should #include the stdlib header if you want to use exit.

I'm getting the error "too few arguments to function 'exit'."

Matticus, I don't understand.

[PYROMANIAC702]

You do not need exit here: you should either structure your code such that returning at the end of main will do, or use return statements.

I haven't learned return statements yet. How would I structure it right?

[laserlight]

I haven't learned return statements yet. How would I structure it right?

Consider the structure:

if x
    if y
        #1
    else
        #2
else
    if z
        #3
    else
        #4

Despite no return statements or exit, at the end of the code, either #1, #2, #3 or #4 would have been executed, and none of the other three.

[Matticus]

I'm getting the error "too few arguments to function 'exit'."

Matticus, I don't understand.

Did you try passing an argument to the function? "exit()" expects an argument, as described in the description I quoted.

If you don't know how to pass arguments to a function, then you have a lot more studying ahead of you.

[PYROMANIAC702]

Did you try passing an argument to the function? "exit()" expects an argument, as described in the description I quoted.

If you don't know how to pass arguments to a function, then you have a lot more studying ahead of you.

No, I do not... As I already stated, I am a programming noob. I'm only in my first class.

[Matticus]

No, I do not... As I already stated, I am a programming noob. I'm only in my first class.

Well, I won't give away the prize, but it will be difficult to give you a clue without practically giving you the answer. Hopefully, this analogy will help you understand.

  printf ("There is no real solution.\n");

  // "printf()" is a function.
  // "There is no real solution.\n" is the argument passed to that function.

[laserlight]

No, I do not... As I already stated, I am a programming noob. I'm only in my first class.

That is fine. I am going to reiterate: the use of exit here is the wrong approach. Consider how to structure your if statements such that you get the result that you want.

The way I see it, if you want to use exit, it is because some error happened that you cannot recover from, so your only option is to end the program there and then. In this case, you are just defining the logic by which you compute the solution(s) to the equation.

You should eventually learn how to call a function and use return statements, but if you use exit as you intend here, I am really going to facepalm.

[stahta01]

I haven't learned return statements yet.

DO NOT try to use exit! If you not NOT know how to use return you have no business using exit().

After you fix your if statements using "==" instead of "=", try doing what laserlight suggested and rewrite your if statement.

Tim S.

[PYROMANIAC702]

That is fine. I am going to reiterate: the use of exit here is the wrong approach. Consider how to structure your if statements such that you get the result that you want.

The way I see it, if you want to use exit, it is because some error happened that you cannot recover from, so your only option is to end the program there and then. In this case, you are just defining the logic by which you compute the solution(s) to the equation.

Ok, I did what you told me too and it "kind of" worked. I can't do a continuous if, else if , else because then it won't work. Here's what I did, which still doesn't work in one occasion.

#include <stdio.h>
#include <math.h>
#include "genlib.h"
#include "simpio.h"

main()

{
      int a, b, c, X, D, x1, x2;
      
      printf ("This program solves a quadratic equation.\n");
      printf ("Enter the value of A: ");
      a = GetInteger();
      printf ("Enter the value of B: ");
      b = GetInteger();
      printf ("Enter the value of C: ");
      c = GetInteger();
      if (a == 0)
      if (b == 0)
      {
            printf ("no solution.");
            }
            else
            {
                X = - c / b;
                printf ("The equation is not quadratic and the solution is: %d\n", X);
                }
      D = b * b - 4 * a * c;
      if (D < 0)
      {
            printf ("There is no real solution.\n");
            }
            else if (D == 0)
            {
                  x1 = - b / 2 * a;
                  printf ("One solution: %d", x1);
                  }
      else
      {            
      x1 = (- b + sqrt(D)) / 2 * a;
      x2 = (- b - sqrt(D)) / 2 * a;
      printf ("Two solutions: %d and %d", x1, x2);
             }
      getchar();
      }

If I enter 0, 3, -6 it gives me two answers: "The equation is not quadratic and the solution is: 2" and "Two solutions : 0 and 0." I have no clue how to fix this.