Size of a 'long' in memory - Java

Question

Reviewed

I am still trying to figure out why long and double in Java consuming 12 bytes instead of 8 bytes in memory

Sure there is wrong assumptions somewhere; please guide me..

I have analyzed it using two ways,

Following is my logic/assumption/source info,

Define a single float or long property in a class and check its memory footprint
Object will carry its own Meta Data worth 12 bytes = [CLASS INFO OF 4 BYTES] + [FLAGS INFO OF 4 BYTES] + [LOCK INFO OF 4 BYTES]
Class size will be Total Memory - Meta Data
Result received is 12 bytes => 24 bytes (Total Memory) - 12 bytes (Meta Data)
JDK: 1.8.0_65; Java HotSpot(TM) 64-Bit Server VM
OS: Windows 8.1
Test program

[ANSWER] Difference of 4 bytes is owing to Padding Applied by JVM. Thanks Andy Turner...

@AndyTurner but inside the `DoubleClass` and `LongClass` there are primitive `long` and `double` variables. — radoh, Mar 17 '16 at 14:19
@radoh `DoubleClass` and `LongClass` has its own meta data worth 12 bytes + 8 byte primitive `long` or `double` + 4 byte padding => 24 bytes total. I was not able to know that from where 4 bytes were coming in picture. Memory padding was the correct answer by Andy .... :) — InBravo, Mar 18 '16 at 02:45

score 2 · Answer 1 · edited May 23 '17 at 12:07

2

In a modern 64-bit JDK, an object has a 12-byte header, padded to a multiple of 8 bytes

The extra 4 bytes is padding to get to a multiple of 8.

edited May 23 '17 at 12:07

Community

answered Mar 17 '16 at 14:21

Andy Turner

score 0 · Answer 2 · answered Mar 17 '16 at 14:29

0

you can check the size of double:

double numDouble=2;
long size=(long)(numDouble*Double.SIZE) / Byte.SIZE;
System.out.println(size);

output: 16

answered Mar 17 '16 at 14:29

Musaddique S