Memory alignment of Java classes

Question

Suppose that I am in a 64-bit machine compiling a C program with gcc. I'm assuming that sizeof(int) is 8 bytes, and sizeof(char) is 1 byte.

Because of memory alignment, the following struct:

struct example{
    int a;
    char c;
}

does not actually have a size of 9 bytes, but 16 (twice sizeof(int)), so that both its beginning and ending addresses can be at multiples of the word size (assumed 8 bytes here).

I was wondering how large the following class would be in Java 8:

class Node {
    int val;
    Node left, right;
    boolean flag;
 }

I'm basically not certain whether we would align at multiples of 8 or 4 bytes.

Answer 1

You can use jol to know the exact layout of objects. This is the output of the program for your Node class (on Oracle JDK 1.8.0_121 64 bit):

# Running 64-bit HotSpot VM.
# Using compressed oop with 3-bit shift.
# Using compressed klass with 3-bit shift.
# Objects are 8 bytes aligned.
# Field sizes by type: 4, 1, 1, 2, 2, 4, 4, 8, 8 [bytes]
# Array element sizes: 4, 1, 1, 2, 2, 4, 4, 8, 8 [bytes]

org.example.Node object internals:
OFFSET  SIZE    TYPE DESCRIPTION                    VALUE
     0     4         (object header)                01 00 00 00 (00000001 00000000 00000000 00000000) (1)
     4     4         (object header)                00 00 00 00 (00000000 00000000 00000000 00000000) (0)
     8     4         (object header)                70 22 01 f8 (01110000 00100010 00000001 11111000) (-134143376)
    12     4     int Node.val                       0
    16     1 boolean Node.flag                      false
    17     3         (alignment/padding gap)        N/A
    20     4    Node Node.left                      null
    24     4    Node Node.right                     null
    28     4         (loss due to the next object alignment)
Instance size: 32 bytes
Space losses: 3 bytes internal + 4 bytes external = 7 bytes total

So, the alignment is 8 bytes.

Note, this is platform-specific. You shouldn't rely much on this information.

Answer 2

ZhekaKozlov's answer is not correct.

In Hotspot JVM on JDK8 at least, the alignment is 8 bytes by default, whether it's 32-bit JVM or 64-bit.

(Sorry I don't have enough reps to comment, so I have to post a new answer )

Tried on Oracle jdk-8u251 x86 version,

You can try starting your (Hotspot) JVM with the option XX:ObjectAlignmentInBytes=4, and it will fail with error:

ObjectAlignmentInBytes=4 must be greater or equal 8

Interesting sidenote, if you set it as 9, it will also complain that the number must be power of 2.

Also run this code snippet on the JVM above without any explicit VM option set, using jol,

System.out.println(VM.current().details());

it will print

# Running 32-bit HotSpot VM.
# Objects are 8 bytes aligned.
# Field sizes by type: 4, 1, 1, 2, 2, 4, 4, 8, 8 [bytes]
# Array element sizes: 4, 1, 1, 2, 2, 4, 4, 8, 8 [bytes]

Do note that the alignment size is not defined in JVM specification so it might vary by implementation.