java programming - dynamic variable names

Question

I had a question regarding this program I was trying to write.

fin = new FileInputStream (args[0]);
            BufferedReader d = new BufferedReader(new InputStreamReader(fin));
            String str;

            while ((str = d.readLine()) != null) {
                String [] string = str.split(" ");
                System.out.println(string[0]);
                int i=0;
                if(!(string[0].compareTo("INPUT")==0) ||  (string[0].compareTo("OUTPUT")==0))
                {
                    // solution goes here

                }
            }

            fin.close();

So the question i have is that i am trying to create objects a1, a2, a3 etc of a class (say A) and then add them as a vertex to a graph. Is there a way to define them dynamically? Would this work?

A [] a;
while(condition is true)
{
 a[i].setParameters() // dont worry about this
 graph.addVertex(a[i])
 i++;
}

If not, can you tell me a better way of doing this? When i try this i get local variable not initialized error. I might be missing something trivial

Note: The project is to design a logic circuit simulator and I am trying to implement a graph styled data structure where nodes depict gates and the interconnects will be edges (I have two classes called gates and interconnects)

Why not try to compile and test first before asking here? You'd be able to ask a much more informative and specific question this way. — Hovercraft Full Of Eels
– Hovercraft Full Of Eels, Commented Oct 16, 2011 at 22:52
@HovercraftFullOfEels I did. I wrote in my question it gives me a local variable not initialized error. I was mentioning my solution so that you guys can know i tried before asking on SO — SP6
– SP6, Commented Oct 16, 2011 at 22:55
Please try to write more descriptive question titles. We know that it's programming (you're posting on SO) and we know that it's Java (from the tag). — Matt Ball
– Matt Ball, Commented Oct 16, 2011 at 22:56
If you need more info regarding my question please let me know which part is not clear. — SP6
– SP6, Commented Oct 16, 2011 at 22:56

b.buchhold · Accepted Answer · 2011-10-16 22:57:55Z

2

It is hard to tell if this fixes all of your problems, since you didn't provide a precise error message. At least it resolves a problem. You never created objects in your array. You'd have to know the since upfornt to do so, or you can achieve the desired behavior just like this:

while(condition is true)
{
 A a = new A();
 a.setParameters(); // dont worry about this
 graph.addVertex(a);
 // If you also need it in some kind of array, add it to the array, too.
 // Might be nice to use Vector<A> or List<A> aList = new ArrayList<A>() 
 // since those two can outmaticcaly grow while inserting at the back.
 // Create those outside of the loop if you need them at all.
 i++;
}

answered Oct 16, 2011 at 22:57

b.buchhold

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2 Comments

Bohemian Over a year ago

My thoughts exactly - don't keep a reference to the value if you don't need to - just send it in to addVertex() then throw it away.

SP6 Over a year ago

No i need the arrays. I think I fixed my problem with list

Brian Roach · Accepted Answer · 2011-10-16 22:56:05Z

1

Problem #1:

A [] a;

This only declares a. You still have to create the array with:

a = new A[5]; // Or at the same time with: A[] a = new A[5];

for example, to get an array that can hold 5 A objects. Then each element needs to be created:

a[0] = new A();
...

Problem #2: As you can see, arrays aren't dynamic, which is problematic if you don't know how large they need to be ahead of time. You need to look at perhaps using a collection, such as an ArrayList

answered Oct 16, 2011 at 22:56

Brian Roach

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Comments

dsolimano · Accepted Answer · 2011-10-16 23:00:39Z

1

It looks like you need to initialize the array a. Instead of

A [] a;

you need

A[] a = new A[100];//whatever size you need.

If you don't know how many As you will be creating, some sort of dynamically expanding structure like a List might be more appropriate.

List<A> listOfA = new ArrayList<A>();
while(condition is true)
{
  A a = createASomehow();
  a.setParameters();
  listOfA.add(a);
  graph.addVertex(a)
}

Though really, I'm not sure quite why you need to keep the set of As on the side, does the graph not have some method that returns a list of its vertices?

answered Oct 16, 2011 at 23:00

dsolimano

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