How to continuously prompt error message in an array

Question

How can I keep prompting the error message until a valid grade has been entered? Right now, I'm only prompted 4 times then the average is displayed.

Also, how would I calculate the average of only the three highest grades?

  final int MAX_NUM_GRADES = 4;
      double avg;
      double sum = 0;
      int count = 1;
      double[] examGrade = new double[MAX_NUM_GRADES];
      for (int i = 0; i < examGrade.length; i++) {
         try {
            examGrade[i] = Double.parseDouble(JOptionPane.showInputDialog("Enter Grade " + count + ":"));
            if (examGrade[i] < 0 || examGrade[i] > 100) {
               JOptionPane.showMessageDialog(null,"ERROR!.");           
            }
            else {
               count++;
               sum = sum + examGrade[i];                       
            }
         }
         catch (NumberFormatException e) {
            JOptionPane.showMessageDialog(null,"ERROR!");         
         }      
      }
      avg = sum/examGrade.length;
      JOptionPane.showMessageDialog(null,"average exam grade is: " 
         + String.format("%.1f",avg));

Use a do-while or a while-do loop instead of a for loop; the termination condition would be eventually "when a valid grade is entered". Notice although you can use something similar with a for loop but the semantics might look "odd" — user4695271
– user4695271, Commented Jun 16, 2015 at 17:46

Balkrishna Rawool · Accepted Answer · 2015-06-16 18:28:13Z

1

Change your code to this:

  final int MAX_NUM_GRADES = 4;
  double avg;
  double sum = 0;
  double min = Double.MAX_VALUE;
  int count = 1;
  double[] examGrade = new double[MAX_NUM_GRADES];
  for (int i = 0; i < examGrade.length; i++) {
     boolean invalid = true;
     while(invalid) {
     try {
        examGrade[i] = Double.parseDouble(JOptionPane.showInputDialog("Enter Grade " + count + ":"));
        if (examGrade[i] < 0 || examGrade[i] > 100) {
           JOptionPane.showMessageDialog(null,"ERROR!.");           
        }
        else {
           invalid = false;
           count++;
           sum = sum + examGrade[i];
           if(examGrade[i] < min) min = examGrade[i]; 
        }
     }
     catch (NumberFormatException e) {
        JOptionPane.showMessageDialog(null,"ERROR!");         
     }
     }      
  }
  avg = (sum - min)/(examGrade.length - 1);
  JOptionPane.showMessageDialog(null,"average exam grade is: " 
     + String.format("%.1f",avg));

edited Jun 16, 2015 at 18:28

answered Jun 16, 2015 at 17:57

Balkrishna Rawool

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2 Comments

dan Over a year ago

Why does double min = Integer.MAX_VALUE?

Balkrishna Rawool Over a year ago

Changed it to Double.MAX_VALUE. It is initialized with the highest value so it can be used to calculate minimum. Initializing it with any other value would affect minimum calculation.

Sh4d0wsPlyr · Accepted Answer · 2015-06-16 17:46:26Z

1

I would recommend putting it in a loop. Below is a psudo-code example.

for(int I = 0; I < examGrade.length; I++) {
    invalid = true;
     while(invalid) {
        //Accept the input, set invalid = false if it is correct
    }
}

answered Jun 16, 2015 at 17:46

Sh4d0wsPlyr

97813 silver badges35 bronze badges

Comments

vlatkozelka · Accepted Answer · 2015-06-16 17:51:34Z

you need a while or a do while loop , you keep looping till the input is valid

final int MAX_NUM_GRADES = 4;
  double avg;
  double sum = 0;
  int count = 1;
  double[] examGrade = new double[MAX_NUM_GRADES];
  boolean valid; // this is true when the input is valid
  for (int i = 0; i < examGrade.length; i++) {
     valid=false;// initialize valid
     try {
        do{
        examGrade[i] = Double.parseDouble(JOptionPane.showInputDialog("Enter Grade " + count + ":"));
        if (examGrade[i] < 0 || examGrade[i] > 100) {
           JOptionPane.showMessageDialog(null,"ERROR!.");
           valid=false;//in this case input is not valid           
        }
        else {
           count++;
           sum = sum + examGrade[i];
           valid=true;// input is valid                       
        }
       }while(!valid)//keep looping while valid is false
     }
     catch (NumberFormatException e) {
        JOptionPane.showMessageDialog(null,"ERROR!");         
     }      
  }
  avg = sum/examGrade.length;
  JOptionPane.showMessageDialog(null,"average exam grade is: " 
     + String.format("%.1f",avg));

now to calculate average of 3 highest grades you need to sort the array , then pick the last 3 grades , you can use any sorting algorithm , for example:

int tmp;
for(int i=0;i<examGrade.length;i++){
  for(int j=i;j<examGrade.length;j++{
  if(examGrade[j]<examGrade[i]){
    tmp=examGrade[j];
    examGrade[j]=examGrade[i];
    examGrade[i]=tmp;
   }
  }
  //now that its sorted you can calculate the average of highest scores
   float avg=0;
   for(int i=1;i<examGrade.length;i++){
     avg+=examGrade[i];
    }
   avg=avg/3;

You can avoid sorting by using a smarter data structure, fyi

ylun · Accepted Answer · 2015-06-16 18:05:35Z

This is because your for loop is iterating thru only the length of your examGrade array, which is set to 4 at the start of your program with MAX_NUM_GRADES.

Instead, you could try keeping track of valid answers and only calculate once 3 valid grades have been answered using a while-loop. The TreeSet used below automagically sorts the values inputted into the set, so sorting is done for you.

Like so:

TreeSet<Double> grades = new TreeSet<Double>();
final int MAX_NUM_GRADES = 4;
double avg;
double sum = 0;
double temp = 0d;

while(grades.size() < MAX_NUM_GRADES){
     temp = Double.parseDouble(JOptionPane.showInputDialog("Enter Grade " + count + ":"));
     if (examGrade[i] < 0 || examGrade[i] > 100) {
         JOptionPane.showMessageDialog(null,"ERROR!.");           
     } else {
         grades.add(temp);                   
     }
}
for (Double val : grades) {
    sum += val;
}
avg = sum/grades.size();
  JOptionPane.showMessageDialog(null,"average exam grade is: " 
     + String.format("%.1f",avg));

Just add the try-catch blocks where you would like them.

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