Trick behind JVM's compressed Oops

Question

So I understand the compressed oops is enabled by default in HotSpot VM now. It has support for this from Java SE 6u23 onwards through the VM option -XX:+UseCompressedOops. I understand that it allows for efficient CPU cache utilization as the CPU caches can hold a larger number of references than if they had to deal with 64 bit sized references. But what I do not understand is how using only 32 bits JVM can address up to 2⁶⁴ addresses.

To simplify the problem how can we address up to 2⁴ memory address's using just 2 bits? What can be a possible encoding/decoding of such an address scheme?

Answer 1

For a detailed explanation of compressed oops, see the "Compressed oops in the Hotspot JVM" article by John Rose @ Oracle.

The TL;DR version is:

• on modern computer architectures, memory addresses are byte addresses,
• Java object references are addresses that point to the start of a word¹,
• on a 64-bit machine, word alignment means that that the bottom 3 bits of an object reference / address are zero²
• so, by shifting an address 3 bits to the right, we can "compress" up to a 35 bits of a 64 bit address into a 32-bit word,
• and, decompression can be done by shifting 3 bits to the left, which puts those 3 zero bits back,
• 35 bits of addressing allows us to represent object pointers for up to 32 GB of heap memory using compressed oops that fit in 32-bit (half-)words on a 64-bit machine.

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Note that this only works on a 64-bit JVM. We still need to be able to address the memory containing that (up to) 32 GB heap¹, and that means 64-bit hardware addresses (on modern CPUs / computer architectures).

Note also that there is a small penalty in doing this; i.e. the shift instructions required to translate between regular and compressed references. However, the flip side is that less actual memory is consumed³, and memory caches are typically more effective as a consequence.

^{1 - This is because modern computer architectures are optimized for word-aligned memory access.}

^{2 - This assumes that you haven't used -XX:ObjectAlignmentInBytes to increase the alignment from its default (and minimum) value of 8 bytes.}

^{3 - In fact, the memory saving is application specific. It depends on the average object alignment wastage, ratios of reference to non-reference fields and so on. It gets more complicated if you consider tuning the object alignment.}

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To simplify the problem how can we address up to 2⁴ memory addresses using just 2 bits? What can be a possible encoding/decoding of such an address scheme?

You can't address 2⁴ byte addresses. But you can address 2² word addresses (assuming 32-bit words) using 2-bit word addresses. If you can assume that all byte addresses are word-aligned, then you can compress a 4-bit byte address as 2-bit word address by shifting it by 2-bit positions.