what's wrong with this code?

[Ameer Alqallaf]

Hi,
I've been trying to write my first code ever..
I think I somehow managed to do it but there is a compiler error and I can't figure out what it is.
The program suppose to ask the user for 5 numers and then it print the maximum value and the minumum value.
Here is what I wrote:

package firstA;
import java.util.Scanner ;
 class FirstA {

  
    public static void main(String[] args) {
        int a, b, c, d, e, max = 0, min  = 0;
        Scanner keyboard = new Scanner(System.in);
              
        System.out.println ("Enter five numbers") ;
        a = keyboard.netxtInt ();
        b = keyboard.netxtInt ();
        c = keyboard.netxtInt ();        
        d = keyboard.netxtInt ();        
        e = keyboard.netxtInt ();
       
        if (a > b && a > c && a > d && a > e)
            max = a ;
        else if (b > a && b > c && b > d && b > e)
            max = b ;
        else if (d > a && d > b && d > c && d > e)
            max = d ;
        else if (c > a && c > b && c > d && c > e)
            max = c ;
        else if (d > a && d > b && d > c && d > e)
            max = d ;
        else if (e > a && e > b && e > c && e > d)
            max = e ;
        System.out.println("the maximum number is" + max);
        
       if (a < b && a < c && a < d && a < e)
            min = a ;
        else if (b < a && b < c && b < d && b < e)
            min = b ;
        else if (d < a && d < b && d < c && d < e)
            min = d ;
        else if (c < a && c < b && c < d && c < e)
            min = c ;
        else if (d < a && d < b && d < c && d < e)
            min = d ;
        else if (e < a && e < b && e < c && e < d)
            min = e ;
        
        System.out.println("the minimum number is" + min);
    }
    
} 

and here is the error I get
Exception in thread "main" java.lang.RuntimeException: Uncompilable source code - Erroneous sym type: java.util.Scanner.netxtInt
Enter five numbers
at firstA.FirstA.main(First1.java:11)

the wrrors are in these lines
a = keyboard.netxtInt ();
b = keyboard.netxtInt ();
c = keyboard.netxtInt ();
d = keyboard.netxtInt ();
e = keyboard.netxtInt ();

PS. I know there maybe another way to do it but I want to do it this way, and please exuse the mess because I am totaly new to this forum.

[Matthew Brown]

Hi Ameer. Welcome to The Ranch!

Check your spelling of "netxtInt".

Edit: A more general comment - I'd suggest using an array of ints instead of five separate variables, as you'll be able to replace your complicated else/if statements with a loop and a simple test. Imagine what happens if you're now asked "that's great, but can you modify it to take 6/7/8 numbers instead?". With an array, that would be a matter of changing one number. With your approach, you've got a lot of changes to make.

[Ameer Alqallaf]

Thank you very much for you fast replay.

it was really a silly mistake
I'm sure it is because I didn't sleep unitl now trying to write this code

thanks for your comment , you are write about the arry thing , and that is why I mentioned in my post that I know there are other methods ut I was asked to do it this way without arrys as I am totaly beggener to programing and I need to learn it step y step :-)

thank you once more and have a good day

[Joe Hepp]

What happens in your code if your highest or lowest values are entered more than once?

Enter 5,1,2,1,5 for example.

Something to think about.

[Ameer Alqallaf]

I've tried that. It gave me the max and min = 0 !
is there a way ti fix that?!

thanks

[Rob Spoor]

I'd start with using >= instead of >. If "a" > "b" then "a" is the max of "a" and "b". But if "a" >= "b" then "a" is still the max of "a" and "b". It's just that "b" might also be the max in the case when "a" == "b".

The easy way is of course to use nested calls to Math.max / Math.min:

max = Math.max(a, Math.max(b, Math.max(c, Math.max(d, e))));
min = Math.min(a, Math.min(b, Math.min(c, Math.min(d, e))));

[Pranav Raulkar]

Hi Ameer,
The solution is very easy, replace every < by <= and > by >= and you will get the correct results for repeating elements.

[Ameer Alqallaf]

Wow, you guys are very amazing and fast )

Thanks for help, now I'm trying to change the code using JOptionpane to get the numbers and show the result in visual box instead of entering it in the counsil
and I'm stuck here

package firstB;
import javax.swing.JOptionPane ;

 class FirstB {

        public static void main(String[] args) {
        int a, b, c, d, e, max = 0, min  = 0;
       
                    
        JOptionPane.showInputDialog("Enter the first number");
        JOptionPane.showInputDialog("Enter the second number");
        JOptionPane.showInputDialog("Enter the third number");
        JOptionPane.showInputDialog("Enter the fourth number");
        JOptionPane.showInputDialog("Enter the fifth number");

any help what to do next ?

[Hunter McMillen]

The user's response is returned from the JOptionPane.showInputDialog() call. You just need to assign it to a variable.

Hunter

[Rob Spoor]

And convert it from a String into an int. java.lang.Integer has some static methods to help you out.

[Ameer Alqallaf]

here is my final project

package firstB;
import javax.swing.JOptionPane ;

 class FirstB {

        public static void main(String[] args) {
        
            String  firstNumber , secondNumber , thirdNumber , 
        fourthNumber , fifthNumber ;
       
        int a , b , c , d , e , max = 0 , min = 0 ;
        
                    
        firstNumber = JOptionPane.showInputDialog("Enter the first number");
        secondNumber = JOptionPane.showInputDialog("Enter the second number");
        thirdNumber = JOptionPane.showInputDialog("Enter the third number");
        fourthNumber = JOptionPane.showInputDialog("Enter the fourth number");
        fifthNumber = JOptionPane.showInputDialog("Enter the fifth number");
        
        a = Integer.parseInt(firstNumber);
        b = Integer.parseInt(secondNumber);
        c = Integer.parseInt(thirdNumber);        
        d = Integer.parseInt(fourthNumber);     
        e = Integer.parseInt(fifthNumber);
        
        if (a >= b && a >= c && a >= d && a >= e)
            max = a ;
        else if (b >= a && b >= c && b >= d && b >= e)
            max = b ;
        else if (d >= a && d >= b && d >= c && d >= e)
            max = d ;
        else if (c >= a && c >= b && c >= d && c >= e)
            max = c ;
        else if (d >= a && d >= b && d >= c && d >= e)
            max = d ;
        else if (e >= a && e >= b && e >= c && e >= d)
            max = e ;
        JOptionPane.showMessageDialog(null, "the maximum number is" + max);
        
       if (a <= b && a <= c && a <= d && a <= e)
            min = a ;
        else if (b <= a && b <= c && b <= d && b <= e)
            min = b ;
        else if (d <= a && d <= b && d <= c && d <= e)
            min = d ;
        else if (c <= a && c <= b && c <= d && c <= e)
            min = c ;
        else if (d <= a && d <= b && d <= c && d <= e)
            min = d ;
        else if (e <= a && e <= b && e <= c && e <= d)
            min = e ;
        
         JOptionPane.showMessageDialog(null, "the minimum number is" + min);
    }
    
}

Is there any modifications you suggest me to do with it ?
Rob thanks for the advice about using math.max it would really make it a lot easier but as I said I'm trying to learn java step by step and I would defenetly use your advice in the future

[Bill foster]

Hello friend,

I worked on the code that was submitted to the forum the first time.
I was able to get this running I did a test case using numbers in this sequence1,2,3,4,5.

Result was Max was 5 Min was 1

Then I reverse the order of the sequence 5,4,3,2,1.

Result was Max was 5 Min was 1.

The last entry I noticed you had two sets of decision blocks. You might want to try using the decision block in the code with the scanner 9from the first version of your code) with the JOptionPane one (the last update to your code) and see if it works. Your algorithm(if/else statements) worked in the first version of your code. See if it works.

You might want to test the code using numbers like 10, 3, 30, 0, and 5 to test the way the program handles out of sequence numbers. Here's the code:

/*

Bill Foster
Forum question from java ranch

*/

import java.util.*;  
 class FirstA {  
  
    
    public static void main(String[] args) {  
        int a, b, c, d, e, max = 0, min  = 0;  
        Scanner keyboard = new Scanner(System.in);  
                
        System.out.println ("Enter five numbers") ;
		  
		  /*
		  Note this syntax .netxtInt is wrong you can view the Scanner class in the API on the Orical web-site
		  to import all of the methods associated with the util (Scanner Class) package (use java.util.*;  to get all of the fuctionality of util package)
        a = keyboard.netxtInt ();  
        b = keyboard.netxtInt ();  
        c = keyboard.netxtInt ();          
        d = keyboard.netxtInt ();          
        e = keyboard.netxtInt ();  
		  */  
    // code above is your code below is the revised your algorithm is right in this version      
		  a = keyboard. nextInt ();
         
		  b = keyboard. nextInt (); 
        
		  c = keyboard. nextInt (); 
 
		  d = keyboard. nextInt ();         
        e = keyboard. nextInt ();          
        
         
        if (a > b && a > c && a > d && a > e)  
            max = a ;  
        else if (b > a && b > c && b > d && b > e)  
            max = b ;  
        else if (d > a && d > b && d > c && d > e)  
            max = d ;  
        else if (c > a && c > b && c > d && c > e)  
            max = c ;  
        else if (d > a && d > b && d > c && d > e)  
            max = d ;  
        else if (e > a && e > b && e > c && e > d)  
            max = e ;  
        System.out.println("the maximum number is" + max);  
          
       if (a < b && a < c && a < d && a < e)  
            min = a ;  
        else if (b < a && b < c && b < d && b < e)  
            min = b ;  
        else if (d < a && d < b && d < c && d < e)  
            min = d ;  
        else if (c < a && c < b && c < d && c < e)  
            min = c ;  
        else if (d < a && d < b && d < c && d < e)  
            min = d ;  
        else if (e < a && e < b && e < c && e < d)  
            min = e ;  
          
        System.out.println("the minimum number is" + min);  
    }  
      
}   

 

Try to trace the process using paper and pencil then when you think you have a understanding then compile and see if you were right. This will make you think like a compiler and you can start reading code. Note I commented out the code at the beginning of the program. Check out the comments I gave you some pointers.



Your friend,

Bill Foster

[Bill foster]

Hi again,

I'm sorry I read the posts and noticed that testing with all numbers of the same value will give you 0.
You might try to check the If/Else statements, I did not catch that.

Bill Foster.

[Bill foster]

Hi friend,

Here is the update code for the alagorithm:

 if (a >= b && a >= c && a >= d && a >= e)  
            max = a ;  
        else if (b >= a && b >= c && b >= d && b >= e)  
            max = b ;  
        else if (d >= a && d >= b && d >= c && d >= e)  
            max = d ;  
        else if (c >= a && c >= b && c >= d && c >= e)  
            max = c ;  
        else if (d >= a && d >= b && d >= c && d >= e)  
            max = d ;  
        else if (e >= a && e >= b && e>= c && e >= d)  
            max = e ;  
        System.out.println("the maximum number is" + max);  
          
       if (a <= b && a <= c && a <= d && a <= e)  
            min = a ;  
        else if (b <= a && b <= c && b <= d && b <= e)  
            min = b ;  
        else if (d <= a && d <= b && d <= c && d <= e)  
            min = d ;  
        else if (c <= a && c <= b && c <= d && c <= e)  
            min = c ;  
        else if (d <= a && d <= b && d <= c && d <= e)  
            min = d ;  
        else if (e <= a && e <= b && e <= c && e <=d)  
            min = e ;  
          
        System.out.println("the minimum number is" + min);  

If you have 1,1,1,1,1. then result is Max = 1 and Min = 1 in words the same.
this passes the test cases for multiply same values. i.e. 10,10,20,30,50. Result is Min=10 and Max is 50.

That was a brain teaser


your friend,

Bill Foster

[Hunter McMillen]

There are simpler ways of finding the max and min of a list of numbers. Perhaps you should consider storing a-e in a structure so you can utilize those.

Hunter

[Bill foster]

My fellowRanchPeeps,

I can do a int type array[] and then use a for {} loop to process the algorithm.

I'll see if I can do that, after some sleep

But if a programmer does not know how to use a container or a for loop the above post will do fine, there is more than one way to skin a cat.

That was a niffy trick with the JOptionPane (Using swing package--forgot about that, should look at the API for JOptionPane).

Right now I'm working with Assignment, Increment and decrement of variables. Doing some self study.

Good Night!

Your friend,

Bill Foster