Why do Java objects have to be a multiple of 8?

Question

I know that Java uses padding; objects have to be a multiple of 8 bytes. However, I dont see the purpose of it. What is it used for? What exactly is its main purpose?

Answer 1

Its purpose is alignment, which allows for faster memory access at the cost of some space. If data is unaligned, then the processor needs to do some shifts to access it after loading the memory.

Additionally, garbage collection is simplified (and sped up) the larger the size of the smallest allocation unit.

It's unlikely that Java has a requirement of 8 bytes (except on 64-bit systems), but since 32-bit architectures were the norm when Java was created it's possible that 4-byte alignment is required in the Java standard.

Answer 2

The accepted answer is speculation (but partially correct). Here is the real answer.

First of, to @U2EF1's credit, one of the benefits of 8-byte boundaries is that 8-bytes are the optimal access on most processors. However, there was more to the decision than that.

If you have 32-bit references, you can address up to 2^32 or 4 GB of memory (practically you get less though, more like 3.5 GB). If you have 64-bit references, you can address 2^64, which is terrabytes of memory. However, with 64-bit references, everything tends to slow down and take more space. This is due to the overhead of 32-bit processors dealing with 64-bits, and on all processors more GC cycles due to less space and more garbage collection.

So, the creators took a middle ground and decided on 35-bit references, which allow up to 2^35 or 32 GB of memory and take up less space so to have the same performance benefits of 32-bit references. This is done by taking a 32-bit reference and shifting it left 3 bits when reading, and shifting it right 3 bits when storing references. That means all objects must be aligned on 2^3 boundaries (8 bytes). These are called compressed ordinary object pointers or compressed oops.

Why not 36-bit references for accessing 64 GB of memory? Well, it was a tradeoff. You'd require a lot of wasted space for 16-byte alignments, and as far as I know the vast majority of processors receive no speed benefit from 16-byte alignments as opposed to 8-byte alignments.

Note that the JVM doesn't bother using compressed oops unless the maximum memory is set to be above 4 GB, which it does not by default. You can actually enable them with the -XX:+UsedCompressedOops flag.

This was back in the day of 32-bit VMs to provide the extra available memory on 64-bit systems. As far as I know, there is no limitation with 64-bit VMs.

Source: Java Performance: The Definitive Guide, Chapter 8

Answer 3

Data type sizes in Java are multiples of 8 bits (not bytes) because word sizes in most modern processors are multiples of 8-bits: 16-bits, 32-bits, 64-bits. In this way a field in an object can be made to fit ("aligned") in a word or words and waste as little space as possible, taking advantage of the underlying processor's instructions for operating on word-sized data.