Thread: Program that determines the lagest and the smallest number

[B]What is Program about?[/B]
 
I am working on this program which will keep getting input from user until 0 is entered. when 0 is entered program 
will show which was the largest and the smallest number of all the numbers entered.
 
[B]What is the error?[/B]
 
Program below works and when I enter 0 it shows the largest value but shows 0 as the smallest value. 
Example: if i enter few values like 5, 10, 15 and than enter 0 to get the results, it shows the correct 
largest number which in this case is 15 but shows 0 as the smallest number which in this case is 5. How can that be corrected? 
 
import java.util.Scanner; 
  public class LargestAndSmallest
    {
    public static void main(String [] args)
    {
    int number; 
    int largeNumber = 0;
    int smallNumber = 0;
 
    Scanner s = new Scanner(System.in);
 
    System.out.println("Enter multiple numbers\n" +
                       "and enter 0 to get the result of\n"+
                       "largest and smallest numbers entered");
 
    System.out.print("Enter a number or 0 to end or get results: ");
    number = s.nextInt();
    while (number != 0)
    {
 
    System.out.print("Enter a number or 0 to end or get results: ");
    number = s.nextInt();
 
    if(number < smallNumber )
      smallNumber = number;
    if (number > largeNumber )
      largeNumber = number;
    }
    System.out.println(smallNumber);
    System.out.println(largeNumber);
    }
    }

Originally Posted by GoodbyeWorld

You code should end immediately if they enter 0 if you want it to stop then. Otherwise, it's going to operate on 0, and make 0 the smallest number because it will be.

Also, I might add that unless the user enters 0 right away, the value of that number that the user entered before the while loop will be immediately overwritten in the while loop.

Also, I might add that since smallNumber is already initialized to 0 even before the loop, that, unless you enter a negative, the smallest value will be 0.

Make smallNumber the largest possible int. (I think it's 2147483647. It can also be obtained by setting smallNumber to Integer.MAX_VALUE)
The same thing could, in theory, happen for largest if you entered

-20 -15 -10 -5 0 (to stop and not counting 0.) It would return 0 as the largest even though the largest was -5.

Make largeNumber the smallest possible int (I think it's -2147483648. It can also be obtained by setting largeNumber to Integer.MIN_VALUE)

If you want it to always end and NEVER count 0, then a simple if/else statement should work to make sure it doesn't run the code on 0 but just exits the while loop.

Is there any way of solving it without these Integer.MIN_VALUE and MIX_VALUE etc?

some one replied me in other thread and told me to do whats written below but I can't figure that out. Can you?

int smallNumber = 0;

You are comparing to this value to find the smallest number, so unless you have some negative numbers, 0 will always be the smallest.

I would set smallNumber to the first value entered by the user, then compare the other numbers to this.

A quick fix would be to initialize smallNumber with a large number />/>/>. That is, a number that will always be larger than those entered by the user.

Originally Posted by Junky

Do you have a question or are you just posting the help you got in another forum? BTW thanks for wasting everyones time if you were getting help elsewhere.

No problem. You are welcome spoon feeders.

Lol was that an insult? Confusing but funny.

Anyway, smallNumber and largeNumber should be initialised or at least set with the first number entered by the user. Not some arbitrary large number. Use the actual data ffs.