While researching about arrays in java, I read the memory usage contains 12 bytes of header object plus the storage of the element of the particular data type. But if the final value is not divisible by 8 bytes padding needs to be added. Why is this? I tried to search about this but did not find an answer. https://study.com/academy/lesson/java-arrays-memory-use-performance.html#:~:text=The%20memory%20allocation%20for%20an,a%20multiple%20of%208%20bytes. This was the website where I read about this.
2 Answers
The details you’re specifying here are specific to particular implementations of the JVM and are not generally required to be true. For example, as a project many years back I put together an implementation of a JVM written purely in JavaScript, where the implementation relied on underlying JS primitives and therefore couldn’t control the exact sizes of the objects being created. I have no idea how big the underlying array objects I allocated in JavaScript to represent a single Java array was, and I certainly didn’t manually pad it. :-)
As @cpvmrd mentioned, a specific implementation of the JVM might decide to pad things this way either for performance reasons (to align objects to a particular boundary to make 64-bit loads fast) or for reasons of processor alignment (some operations either lead to performance degradation or to bus errors if data of size N are loaded from an address that isn’t a multiple of N). But that’s up to the implementation to sort out.
It is for performance.
The processor reads the memory in "chunks" of a definite size (word). On 64-bits CPU, memory is generally configured to return one 64-bits word per address access. Intel CPUs can perform accesses on non-word boundary，but there is a performance penalty as internally the CPU performs two memory accesses and a math operation to load one word.
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