Java output error

Question

I can't get this to work properly. It functions as it is supposed to, and does the math, but then it loops once, and ends. I need it to either loop until the users decides to end it, or only run once.

import java.util.Scanner;

public class java {
  public static void main(String args[]) {
    System.out.println("Welcome to the simple Calculator program");
    System.out.println("Please type what type of math you would like to do: ");
    System.out.println("1=Addition");
    System.out.println("2=Subtraction");
    System.out.println("3=Multiplication");
    System.out.println("4=Division");
    System.out.println("5=Sqrt");
    Scanner input = new Scanner(System.in);
    int math = input.nextInt();
    if (math == 1) {
      Scanner a = new Scanner(System.in);
      int a1;
      int a2;
      int asum;
      System.out.print("Please enter the first number:  ");
      a1 = a.nextInt();

      System.out.print("Please enter the second number:  ");
      a2 = a.nextInt();

      asum = a2 + a1;
      System.out.print("The sum is: " + asum + "Thank You for using this program");
    }

    Scanner number = new Scanner(System.in);
    int number1;
    int number2;
    int sum;
    System.out.print("Enter first number:  ");
    number1 = number.nextInt();

    System.out.print("Enter Second number:  ");
    number2 = number.nextInt();

    sum = number1 + number2;

    System.out.printf("Sum is %d\n", sum);
  }
}

Why on earth do you reopen a Scanner for each token you are parsing? — fge
– fge, Commented Dec 27, 2012 at 19:20
@user1644742, you need to begin with some more simple, make a loop while(inp != 'q') for example, inside de loop read in var inp with a Scanner(only one) and print the input, you will see when a == 'q' the application will end, in other case will be iterating forever. — HMarioD
– HMarioD, Commented Dec 27, 2012 at 19:32

Smit · Accepted Answer · 2012-12-27 19:25:15Z

1

Use

   do{
      // do something.
   } while(some condition);

And reapeat the same scanner to get input. You can also add one more option to your menu for repeating and evaluate that with while condition.

answered Dec 27, 2012 at 19:25

Smit

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Comments

Abhishek kumar · Accepted Answer · 2012-12-27 19:32:52Z

1

It is working as it should. If you want it to loop as per some user input,you must use any looping construct like while. Instead of if (math == 1) use

`while (math != exit)`

Make a new entry for exit like 0

edited Dec 27, 2012 at 19:32

answered Dec 27, 2012 at 19:24

Abhishek kumar

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Comments

bane · Accepted Answer · 2012-12-27 19:43:16Z

Try using while loop. Give the user an option to quit the program.

import java.util.Scanner;
public class java
{
    public static void main(String args[])
    {
        Scanner a = new Scanner(System.in);
        System.out.println("Welcome to the simple Calculator program");
        while(true)
        {
            System.out.println("Please type what type of math you would like to do: ");
            System.out.println("1=Addition");
            System.out.println("2=Subtraction");
            System.out.println("3=Multiplication");
            System.out.println("4=Division");
            System.out.println("5=Sqrt");
            System.out.println("6=Quit"); // added an option to quit the program
            int math = a.nextInt();
            if (math == 1)
            {
                int a1,a2,asum;
                System.out.print("Please enter the first number:  ");
                a1 = a.nextInt();
                System.out.print("Please enter the second number:  ");
                a2 = a.nextInt();
                asum = a2 + a1;
                System.out.println("The sum is: " + asum + "Thank You for using this program");
            }

            // Include actions for math = 2 to 5

            if(math == 6)
            {
                System.out.println("Thank You for using this program");
                System.exit(0);
            }
        }
    }
}

The options are displayed again and again after each calculation until the user wants to exit the program by entering 6.
If you want the program to run only once, you should leave out the outer while loop. Everything else remains the same.
PS - You don't need to reopen Scanner again and again (at least not in this problem).

ashutosh raina · Accepted Answer · 2012-12-27 19:25:38Z

0

That is because you are only reading the input from the console once..you need to keep the console up with something like while(true) {} or monitor the console for an exit conditon like (" 0 = exit ") .

Also, I don''t think you will need to read two numbers again and again like you are doing right now.

answered Dec 27, 2012 at 19:25

ashutosh raina

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Comments

user1846747 · Accepted Answer · 2012-12-27 19:43:37Z

0

1) You can use a do-while loop with a condition till which you wish to execute.

2) Use a switch case and perform the math operations inside the switch case with different operators. As of now you are trying to perform only addition. So you a switch case where you can perform all the operations.

3) In the switch case have an option which calls the exit(0) method. So that you can run the program until the user wish to exit.

4) By using a switch case you can make the user to choose his own option.

answered Dec 27, 2012 at 19:43

user1846747

Comments

user3636934 · Accepted Answer · 2014-05-20 10:55:00Z

0

Your entire program is correct dude.

Just add

System.exit(0);

In every if(math==1) ,if(math==2)...before their statement ending.

like if(math==1)

{

...

System.exit(0);

}

You can fix your error...

Like me if your error is fixed. If not tell me the error

answered May 20, 2014 at 10:55

user3636934

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