I'm curious about how pointer compression works, especially in the context of Java. I've come across a couple of interesting methods to fit 64-bit pointers into 32-bit values. One approach involves storing offsets from the heap, which is discussed in a V8 blog post, and another method suggests shifting pointers to the right, as outlined in an article by Shipilev. However, I'm left wondering how these techniques allow us to overcome the 4GB heap size limitation. Can anyone clarify this for me?
5 Answers
When pointers are right-shifted, it effectively doubles the limit on what can be accessed, making memory management more efficient in a compressed state.
In Java, references are akin to pointers with some special rules. They typically point to a single area of memory known as the Java heap. Every Java object has a header that takes up space, leading to some complicating factors. This structured nature allows for the omission of certain bits, which in turn enables clever pointer compression tricks.
It looks like Shipilev's article you linked really explains this well. It might be a good idea to re-read it if you're seeking clarity on how pointer compression allows for larger heap sizes.
Pointer compression can work up to 32GB by effectively saving bits. For instance, if you're saving three bits (given the 8-byte distance), the compressed pointers can manage within that limit, letting you tackle the 4GB problem better when needed.
Actually, there’s no hard limitation. The JVM can revert to using 64-bit pointers when necessary, allowing the heap to grow as much as it requires. On 64-bit systems, addresses are aligned to multiples of 8, meaning the last three bits are always zero. This lets the JVM utilize 35-bit addresses, utilizing the three bits in front to cover heaps up to about 32GB efficiently.
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