[japanir]

why use parrallel arrays?
better create a Student class, and have a List of Student Objects, as the instructions require.
the Student Constructor will get a name and the final score:

public class Student {
   private String name;
   private int score;
   
   public Student(String name, int score){
       this.name = name;
       this.score = score;
   }
   //more methods...

then, you can add as much Students to a List and iterate the List to get average score, and all other data you need.

[japanir]

I see.. if you cannot use classes, than you have to use parrallel arrays as you said.
first think if you know the size of the arrays. if not, use a List implementation.
just fill the arrays or Lists (depand on what you choose), and use methods to perform the functions of the instruction.
However, it is really hard to say without seeing some code you wrote (to see your line of thought)

[macosxnerd101]

@Elizabeth: I've written a tutorial on class design called Moving Away From Parallel Arrays. In it, I explain why classes are a better choice than parallel arrays, and how to design and use them in a simple inventory program. If you are required to sort the Students, then using an array of Student objects will be much easier than using parallel arrays, as you can make use of the Arrays.sort() method. Just make sure that your Student class implements the Comparable<Student> interface. Try to use the Object-Oriented model vs. the parallel array model. We will be happy to help you with the code you write.

@japanir: Parallel arrays are one thing, but parallel Lists (or other dynamic collections) are even worse practice, as it is very easy to mismatch the data due to the dynamic resizing.

This post has been edited by macosxnerd101: 10 April 2010 - 11:56 AM

[macosxnerd101]

If you know that name[0] correlates to score[0], you can use a for loop, printing out name[i] and score[i] in the body of the for loop.

[macosxnerd101]

Are you sure? Below, you seem to have no problems using a for loop to access corresponding elements in parallel arrays. If you want to work with a JOptionPane, you can concatenate this data into a single String object to display after the loop. Hint- you may find the \n escape sequence helpful, as it adds a line break.

 for(int k = 0; k < num; k++) 
        { 
                String inputname = JOptionPane.showInputDialog("Name of Student"); 
                name[k] = inputname; 
                int inputscore = Integer.parseInt(JOptionPane.showInputDialog("Score of Student")); 
                score[k] = inputscore; 
        } 

[erik.price]

@macosxnerd101 - I don't think the problem is with printing it out, it is with sorting the one array, and keeping the other matched up.

@Elizabeth - In your sort method, when you swap two elements in your score array, make sure to swap the same two indexes in your name array. So if you end up having to swap score[0] and score[1], you also need to swap name[0] and name[1]

[macosxnerd101]

Quote

How do display a list stored my score with the students name.

@erik.price: That was the question I was going on.

@Elizabeth: There are tons of examples of sorting algorithms in our Java Snippets section. I have a feeling that Selection Sort may be the easiest to understand and implement, though you may also want to look at Bubblesort (least efficient) and possibly Insertion Sort (most efficient out of these three).

[Luckless]

Well, what you should do is, whenever you switch the position of two indexes in list[] (e.g. switch list[1] with list[4]) you should also switch the position of the name[] (e.g. switch name[1] with name[4]). Basically, this would allow you to set up a for loop that prints out list[i] then name[i] and they would all be lined up in the proper order, meaning the name of the student would be in the same position as their score in list[]. Did that make sense?