Java collections: What happens when "size" exceeds "int"?

Question

I know that Java collections are very memory-hungry, and did a test myself, proving that 4GB is barely enough to store few millions of Integers into a HashSet.

But what if I has "enough" memory? What would happen to Collection.size()?

EDIT: Solved: Collection.size() returns Integer.MAX when the integer range is exceeded.
New question: how to determine the "real" count of elements of a collection then?

NOTE 1: Sorry, this is probably a let-me-google-it-for-you question, but I really didn't find anything ;)

NOTE 2: As far as I understand it, each integer entry of a set is: reference + cached_hashcode + boxed_integer_object + real_int_value, right?

NOTE 3: Funny, even with JDK7 and "compressed pointers", when the JVM uses 2GB of real memory, it shows only 1.5GB allocated memory in VisualVM.

For those who care:

Test sources:

import java.util.*;
import java.lang.management.*;

public final class _BoxedValuesInSetMemoryConsumption {
  private final static int MILLION = 1000 * 1000;

  public static void main(String... args) {
    Set<Integer> set = new HashSet<Integer>();

    for (int i = 1;; ++i) {
      if ((i % MILLION) == 0) {
        int milsOfEntries = (i / MILLION);
        long mbytes = ManagementFactory.getMemoryMXBean().
            getHeapMemoryUsage().getUsed() / MILLION;
        int ratio = (int) mbytes / milsOfEntries;
        System.out.println(milsOfEntries + " mil, " + mbytes + " MB used, "
            + " ratio of bytes per entry: " + ratio);
      }

      set.add(i);
    }
  }
}

Execution parameters:

Tested with x64 version of JDK7 build 105 under OpenSuse 11.3 x64.

-XX:+UseCompressedOops -Xmx2048m

Output result:

1 mil, 56 MB used,  ratio of bytes per entry: 56
2 mil, 113 MB used,  ratio of bytes per entry: 56
3 mil, 161 MB used,  ratio of bytes per entry: 53
4 mil, 225 MB used,  ratio of bytes per entry: 56
5 mil, 274 MB used,  ratio of bytes per entry: 54
6 mil, 322 MB used,  ratio of bytes per entry: 53
7 mil, 403 MB used,  ratio of bytes per entry: 57
8 mil, 452 MB used,  ratio of bytes per entry: 56
9 mil, 499 MB used,  ratio of bytes per entry: 55
10 mil, 548 MB used,  ratio of bytes per entry: 54
11 mil, 596 MB used,  ratio of bytes per entry: 54
12 mil, 644 MB used,  ratio of bytes per entry: 53
13 mil, 827 MB used,  ratio of bytes per entry: 63
14 mil, 874 MB used,  ratio of bytes per entry: 62
15 mil, 855 MB used,  ratio of bytes per entry: 57
16 mil, 902 MB used,  ratio of bytes per entry: 56
17 mil, 951 MB used,  ratio of bytes per entry: 55
18 mil, 999 MB used,  ratio of bytes per entry: 55
19 mil, 1047 MB used,  ratio of bytes per entry: 55
20 mil, 1096 MB used,  ratio of bytes per entry: 54
21 mil, 1143 MB used,  ratio of bytes per entry: 54
22 mil, 1191 MB used,  ratio of bytes per entry: 54
23 mil, 1239 MB used,  ratio of bytes per entry: 53
24 mil, 1288 MB used,  ratio of bytes per entry: 53
25 mil, 1337 MB used,  ratio of bytes per entry: 53
Exception in thread "main" java.lang.OutOfMemoryError: Java heap space

At the end, about 2 GiB real memory were used, instead of displayed 1.3 GiB, so the consumption for each entry is even larger than 53 bytes.

Answer 1

I know that Java collections are very memory-hungry, and did a test myself, proving that 4GB is barely enough to store few millions of Integers into a HashSet.

Java Heap != System Memory. Java's default heap size is only 128MB. Note this is also different from the memory the JVM uses.

Regarding your question: from the docs,

public int size()

Returns the number of elements in this collection. If this collection contains more than Integer.MAX_VALUE elements, returns Integer.MAX_VALUE.

Answer 2

Your question seems to have a quite different content than the title.

You already answered the question in the title (Integer.MAX_VALUE is returned). And no: there's no way you can find out the "true" size with the normal APIs safe for iterating over the collection and counting (using a long of course).

If you want to store a Set of int values and you know that the range and amount of values can become very big, then a BitSet might actually be a better implementation:

import java.util.*;
import java.lang.management.*;

public final class IntegersInBitSetMemoryConsumption {
  private final static int MILLION = 1000 * 1000;

  public static void main(String... args) {
    BitSet set = new BitSet(Integer.MAX_VALUE);

    for (int i = 1;; ++i) {
      if ((i % MILLION) == 0) {
        int milsOfEntries = (i / MILLION);
        long mbytes = ManagementFactory.getMemoryMXBean().
            getHeapMemoryUsage().getUsed() / MILLION;
        double ratio = mbytes / milsOfEntries;
        System.out.println(milsOfEntries + " mil, " + mbytes + " MiB used, "
            + " ratio of bytes per entry: " + ratio);
      }

      set.set(i);
    }
  }
}

This will produce a constant-size data structure that can hold all values inside the range without changing size and occupying a relatively small amount of memory (1 bit per possible value plus some overhead).

This method has two drawbacks, however:

• it doesn't support negative int values
• it doesn't provide the Set API

Both can easily be worked around by writing a wrapper that uses two BitSet objects (possibly lazily allocated) to hold the positive and negative value range respectively and implements adapter methods for the Set interface.

Answer 3

From the source code:

 /**
 * Returns the number of elements in this collection.  If this collection
 * contains more than <tt>Integer.MAX_VALUE</tt> elements, returns
 * <tt>Integer.MAX_VALUE</tt>.
 * 
 * @return the number of elements in this collection
 */
int size();