Java – In Java, what is the best way to determine the size of an object

javamemory

I have an application that reads a CSV file with piles of data rows. I give the user a summary of the number of rows based on types of data, but I want to make sure that I don't read in too many rows of data and cause OutOfMemoryErrors. Each row translates into an object. Is there an easy way to find out the size of that object programmatically? Is there a reference that defines how large primitive types and object references are for a VM?

Right now, I have code that says read up to 32,000 rows, but I'd also like to have code that says read as many rows as possible until I've used 32MB of memory. Maybe that is a different question, but I'd still like to know.

Best Answer

You can use the java.lang.instrument package.

Compile and put this class in a JAR:

import java.lang.instrument.Instrumentation;

public class ObjectSizeFetcher {
    private static Instrumentation instrumentation;

    public static void premain(String args, Instrumentation inst) {
        instrumentation = inst;
    }

    public static long getObjectSize(Object o) {
        return instrumentation.getObjectSize(o);
    }
}

Add the following to your MANIFEST.MF:

Premain-Class: ObjectSizeFetcher

Use the getObjectSize() method:

public class C {
    private int x;
    private int y;

    public static void main(String [] args) {
        System.out.println(ObjectSizeFetcher.getObjectSize(new C()));
    }
}

Invoke with:

java -javaagent:ObjectSizeFetcherAgent.jar C

Related Solutions

How to determine the size of the array in C

Executive summary:

int a[17];
size_t n = sizeof(a)/sizeof(a[0]);

Full answer:

To determine the size of your array in bytes, you can use the sizeof operator:

int a[17];
size_t n = sizeof(a);

On my computer, ints are 4 bytes long, so n is 68.

To determine the number of elements in the array, we can divide the total size of the array by the size of the array element. You could do this with the type, like this:

int a[17];
size_t n = sizeof(a) / sizeof(int);

and get the proper answer (68 / 4 = 17), but if the type of a changed you would have a nasty bug if you forgot to change the sizeof(int) as well.

So the preferred divisor is sizeof(a[0]) or the equivalent sizeof(*a), the size of the first element of the array.

int a[17];
size_t n = sizeof(a) / sizeof(a[0]);

Another advantage is that you can now easily parameterize the array name in a macro and get:

#define NELEMS(x)  (sizeof(x) / sizeof((x)[0]))

int a[17];
size_t n = NELEMS(a);

Java – What are the differences between a HashMap and a Hashtable in Java

There are several differences between HashMap and Hashtable in Java:

Hashtable is synchronized, whereas HashMap is not. This makes HashMap better for non-threaded applications, as unsynchronized Objects typically perform better than synchronized ones.
Hashtable does not allow null keys or values. HashMap allows one null key and any number of null values.
One of HashMap's subclasses is LinkedHashMap, so in the event that you'd want predictable iteration order (which is insertion order by default), you could easily swap out the HashMap for a LinkedHashMap. This wouldn't be as easy if you were using Hashtable.

Since synchronization is not an issue for you, I'd recommend HashMap. If synchronization becomes an issue, you may also look at ConcurrentHashMap.

Best Answer

Related Solutions

How to determine the size of the array in C

Java – What are the differences between a HashMap and a Hashtable in Java

Related Topic