Java generic object with varying types.

Question

Imagine having 2 Generic ArrayLists each storing different types. My program will be using only one of these ArrayLists at a time. Is it possible to create a general ArrayList (currentArrayList) which can store both ArrayLists, and use the two ArrayLists without casting.

ArrayList<Integer> arrInt = new ArrayList<>();
arrInt.add(10);
ArrayList<String> arrString = new ArrayList<>();
arrString.add("ten");

ArrayList<XXX> currentArrayList = arrInt;
Integer i = currentArrayList.get(0);

currentArrayList = arrString;
String str = currentArrayList.get(0);

Thanks.

@Azodious how would that work without casting? I don't see how this could be done without having to cast the Strings and Integers after getting them from the list. — eis
– eis, Commented Oct 19, 2012 at 9:08
I would like to implement something similar to the above scenario, since I am creating a program where the application and the user take turns (a conversation) and the application has a script which it needs to follow, and I need to keep track of the latest position the application is in — PB_MLT
– PB_MLT, Commented Oct 19, 2012 at 9:25

Daniel Earwicker · Accepted Answer · 2012-10-19 09:21:53Z

1

The problem is the part where you want to:

use the two ArrayLists without casting.

Definitely not going to happen. The whole point of a static type system is to stop you from being able to treat a general type as a specific type without explicitly asking to do so.

So for example, you could say:

ArrayList<?> currentArrayList = arrInt;
Object i = currentArrayList.get(0);

But notice that i is just an Object, so the compiler is happy - we don't know anything about the things in the array except that they must be Object-based because this is a Java program. But you want to say:

ArrayList<?> currentArrayList = arrInt;
Integer i = currentArrayList.get(0);

It's that second line that isn't ever going to happen without a cast.

UPDATE Question from comments:

Can't the compiler easily infer the type of the ArrayList by looking at the generic type of arrInt ?

Suppose it did that for us. What should it then do with the type of currentArrayList when it sees the line:

currentArrayList = arrString;

Your code as it stands assumes that the compiler is not going to do that kind of inference. Suppose you comment-out that second assignment. Now the compiler could make the inference you suggest, and allow your code to compile. But then in the future if you put back the second assignment, the rest of your code would stop compiling! This would be needlessly confusing.

edited Oct 19, 2012 at 9:21

answered Oct 19, 2012 at 9:08

Daniel Earwicker

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1 Comment

PB_MLT Over a year ago

Can't the compiler easily infer the type of the ArrayList<?> by looking at the generic type of arrInt ?

Leemax · Accepted Answer · 2012-10-19 09:17:28Z

You have to determine the type of the object anyway eventually.

i guess you want to avoid "supresswarning("casting")" kind of warning?

if ture, here it goes:

  ArrayList<Object> list = new ArrayList<Object>();
  list.add(1);
  list.add("abc");
  list.add(new Double(3.1415926));
  list.add(true);

  for (Object o : list) {
     // or more accurate
     String type;
     if (o instanceof Integer) {
        type = "int";
        // logic here...
        int i = (Integer) o; // no casting warning
     } else if (o instanceof String) {
        type = "string";
        // logic here...
     }
     // other possibilities if you want
     else { // unmatched
        type = o.getClass().getSimpleName();
     }
     System.out.println(type + ": " + o);
  }

dngfng · Accepted Answer · 2012-10-19 09:11:41Z

0

Misunderstood the question - why on earth would you want to do that? You are not creating a new ArrayList you are just using a Pointer without the type - its still the same ArrayList!

ArrayList<ArrayList<Object>> array = new ArrayList<ArrayList<Object>>();
array.add(new ArrayList<Integer>());
array.add(new ArrayList<String>());

This should do the trick, since all other possible Objects will inherit from Object in hence be accepted.

edited Oct 19, 2012 at 9:11

answered Oct 19, 2012 at 9:05

dngfng

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1 Comment

Brian Agnew Over a year ago

I don't think he wants a list of lists

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