Size of a byte in memory - Java

Okay, there's been a lot of discussion and not a lot of code :)

Here's a quick benchmark. It's got the normal caveats when it comes to this kind of thing - testing memory has oddities due to JITting etc, but with suitably large numbers it's useful anyway. It has two types, each with 80 members - LotsOfBytes has 80 bytes, LotsOfInts has 80 ints. We build lots of them, make sure they're not GC'd, and check memory usage:

class LotsOfBytes
{
    byte a0, a1, a2, a3, a4, a5, a6, a7, a8, a9, aa, ab, ac, ad, ae, af;
    byte b0, b1, b2, b3, b4, b5, b6, b7, b8, b9, ba, bb, bc, bd, be, bf;
    byte c0, c1, c2, c3, c4, c5, c6, c7, c8, c9, ca, cb, cc, cd, ce, cf;
    byte d0, d1, d2, d3, d4, d5, d6, d7, d8, d9, da, db, dc, dd, de, df;
    byte e0, e1, e2, e3, e4, e5, e6, e7, e8, e9, ea, eb, ec, ed, ee, ef;
}

class LotsOfInts
{
    int a0, a1, a2, a3, a4, a5, a6, a7, a8, a9, aa, ab, ac, ad, ae, af;
    int b0, b1, b2, b3, b4, b5, b6, b7, b8, b9, ba, bb, bc, bd, be, bf;
    int c0, c1, c2, c3, c4, c5, c6, c7, c8, c9, ca, cb, cc, cd, ce, cf;
    int d0, d1, d2, d3, d4, d5, d6, d7, d8, d9, da, db, dc, dd, de, df;
    int e0, e1, e2, e3, e4, e5, e6, e7, e8, e9, ea, eb, ec, ed, ee, ef;
}


public class Test
{
    private static final int SIZE = 1000000;

    public static void main(String[] args) throws Exception
    {        
        LotsOfBytes[] first = new LotsOfBytes[SIZE];
        LotsOfInts[] second = new LotsOfInts[SIZE];

        System.gc();
        long startMem = getMemory();

        for (int i=0; i < SIZE; i++)
        {
            first[i] = new LotsOfBytes();
        }

        System.gc();
        long endMem = getMemory();

        System.out.println ("Size for LotsOfBytes: " + (endMem-startMem));
        System.out.println ("Average size: " + ((endMem-startMem) / ((double)SIZE)));

        System.gc();
        startMem = getMemory();
        for (int i=0; i < SIZE; i++)
        {
            second[i] = new LotsOfInts();
        }
        System.gc();
        endMem = getMemory();

        System.out.println ("Size for LotsOfInts: " + (endMem-startMem));
        System.out.println ("Average size: " + ((endMem-startMem) / ((double)SIZE)));

        // Make sure nothing gets collected
        long total = 0;
        for (int i=0; i < SIZE; i++)
        {
            total += first[i].a0 + second[i].a0;
        }
        System.out.println(total);
    }

    private static long getMemory()
    {
        Runtime runtime = Runtime.getRuntime();
        return runtime.totalMemory() - runtime.freeMemory();
    }
}

Output on my box:

Size for LotsOfBytes: 88811688
Average size: 88.811688
Size for LotsOfInts: 327076360
Average size: 327.07636
0

So obviously there's some overhead - 8 bytes by the looks of it, although somehow only 7 for LotsOfInts (? like I said, there are oddities here) - but the point is that the byte fields appear to be packed in for LotsOfBytes such that it takes (after overhead removal) only a quarter as much memory as LotsOfInts.

Java is never implementation or platform specific (at least as far as primitive type sizes are concerned). They primitive types are always guaranteed to stay the same no matter what platform you're on. This differs from (and was considered an improvement on) C and C++, where some of the primitive types were platform specific.

Since it's faster for the underlying operating system to address four (or eight, in a 64-bit system) bytes at a time, the JVM may allocate more bytes to store a primitive byte, but you can still only store values from -128 to 127 in it.

A revealing exercise is to run javap on some code that does simple things with bytes and ints. You'll see bytecodes that expect int parameters operating on bytes, and bytecodes being inserted to co-erce from one to another.

Note though that arrays of bytes are not stored as arrays of 4-byte values, so a 1024-length byte array will use 1k of memory (Ignoring any overheads).

Yes, a byte variable in Java is in fact 4 bytes in memory. However this doesn't hold true for arrays. The storage of a byte array of 20 bytes is in fact only 20 bytes in memory.

That is because the Java Bytecode Language only knows two integer number types: ints and longs. So it must handle all numbers internally as either type and these types are 4 and 8 bytes in memory.

However, Java knows arrays with every integer number format. So the storage of short arrays is in fact two bytes per entry and one byte per entry for byte arrays.

The reason why I keep saying "the storage of" is that an array is also an object in Java and every object requires multiple bytes of storage on its own, regardless of the storage that instance variables or the array storage in case of arrays require.