1. Overview
In this quick article, we’re going to see what is the footprint of a boolean value in the JVM in different circumstances.
First, we’ll inspect the JVM to see the object sizes. Then, we’ll understand the rationale behind those sizes.
2. Setup
To inspect the memory layout of objects in the JVM, we’re going to use the Java Object Layout (JOL) extensively. Therefore, we need to add the jol-core dependency:
<dependency>
<groupId>org.openjdk.jol</groupId>
<artifactId>jol-core</artifactId>
<version>0.10</version>
</dependency>
3. Object Sizes
If we ask JOL to print the VM details in terms of Object Sizes:
System.out.println(VM.current().details());
When the compressed references are enabled (the default behavior), we’ll see the output:
# 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]
In the first few lines, we can see some general information about the VM. After that, we learn about object sizes:
- Java references consume 4 bytes, booleans/bytes are 1 byte, chars/shorts are 2 bytes, ints/floats are 4 bytes, and finally, longs/doubles are 8 bytes
- These types consume the same amount of memory even when we use them as array elements
So, in the presence of compressed references, each boolean value takes 1 byte. Similarly, each boolean in a boolean[] consumes 1 byte. However, alignment paddings and object headers can increase the space consumed by boolean and boolean[] as we’ll see later.
3.1. No Compressed References
Even if we disable the compressed references via -XX:-UseCompressedOops, the boolean size won’t change at all:
# Field sizes by type: 8, 1, 1, 2, 2, 4, 4, 8, 8 [bytes]
# Array element sizes: 8, 1, 1, 2, 2, 4, 4, 8, 8 [bytes]
On the other hand, Java references are taking twice the memory.
So despite what we might expect at first, booleans are consuming 1 byte instead of just 1 bit.
3.2. Word Tearing
In most architecture, there is no way to access a single bit atomically. Even if we wanted to do so, we probably would end up writing to adjacent bits while updating another one.
One of the design goals of the JVM is to prevent this phenomenon, known as word tearing. That is, in the JVM, every field and array element should be distinct; updates to one field or element must not interact with reads or updates of any other field or element.
*To recap, addressability issues and word tearing are the main reasons why booleans are more than just one single bit.*
4. Ordinary Object Pointers (OOPs)
Now that we know booleans are 1 byte, let’s consider this simple class:
class BooleanWrapper {
private boolean value;
}
If we inspect the memory layout of this class using JOL:
System.out.println(ClassLayout.parseClass(BooleanWrapper.class).toPrintable());
Then JOL will print the memory layout:
OFFSET SIZE TYPE DESCRIPTION VALUE
0 12 (object header) N/A
12 1 boolean BooleanWrapper.value N/A
13 3 (loss due to the next object alignment)
Instance size: 16 bytes
Space losses: 0 bytes internal + 3 bytes external = 3 bytes total
The BooleanWrapper layout consists of:
- 12 bytes for the header, including two mark words and one klass word. The HotSpot JVM uses the mark word to store the GC metadata, identity hashcode and locking information. Also, it uses the klass word to store class metadata such as runtime type checks
- 1 byte for the actual boolean value
- 3 bytes of padding for alignment purposes
By default, object references should be aligned by 8 bytes. Therefore, the JVM adds 3 bytes to 13 bytes of header and boolean to make it 16 bytes.
Therefore, boolean fields may consume more memory because of their field alignment.
4.1. Custom Alignment
If we change the alignment value to 32 via -XX:ObjectAlignmentInBytes=32, then the same class layout changes to:
OFFSET SIZE TYPE DESCRIPTION VALUE
0 12 (object header) N/A
12 1 boolean BooleanWrapper.value N/A
13 19 (loss due to the next object alignment)
Instance size: 32 bytes
Space losses: 0 bytes internal + 19 bytes external = 19 bytes total
As shown above, the JVM adds 19 bytes of padding to make the object size a multiple of 32.
5. Array OOPs
Let’s see how the JVM lays out a boolean array in memory:
boolean[] value = new boolean[3];
System.out.println(ClassLayout.parseInstance(value).toPrintable());
This will print the instance layout as following:
OFFSET SIZE TYPE DESCRIPTION
0 4 (object header) # mark word
4 4 (object header) # mark word
8 4 (object header) # klass word
12 4 (object header) # array length
16 3 boolean [Z.<elements> # [Z means boolean array
19 5 (loss due to the next object alignment)
In addition to two mark words and one klass word, array pointers contain an extra 4 bytes to store their lengths.
Since our array has three elements, the size of the array elements is 3 bytes. However, these 3 bytes will be padded by 5 field alignment bytes to ensure proper alignment.
Although each boolean element in an array is just 1 byte, the whole array consumes much more memory. In other words, we should consider the header and padding overhead while computing the array size.
6. Conclusion
In this quick tutorial, we saw that boolean fields are consuming 1 byte. Also, we learned that we should consider the header and padding overheads in object sizes.
For a more detailed discussion, it’s highly recommended to check out the oops section of the JVM source code. Also, Aleksey Shipilëv has a much more in-depth article in this area.
As usual, all the examples are available over on GitHub.