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In Java, the character data type, char, is represented with 2 bytes. The array of n characters, char[], is represented with 2n+24 bytes.

In general, there is an overhead of 24 bytes for storing an array of n objects (at least if the objects are of primitive type).

Why do we need these additional 24 bytes? How are they used?

EDIT (July 2nd 2015). It was brought to my attention in a comment that an answer to this question is offered here on the programmers StackExchange.

Community
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blazs
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    Answer is [here](http://programmers.stackexchange.com/questions/162546/why-the-overhead-when-allocating-objects-arrays-in-java) – Ajith John Jul 02 '15 at 10:39
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    The length of the array is probably going to be part of it (which allows java not to read into the junk after the array by accident - heartbleed bug anyone?) – Richard Tingle Jul 02 '15 at 10:41

2 Answers2

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It is the object header, it includes information about the object itself (locked bits, marked bits for the GC), a pointer to its class object and the length.

loonytune
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In addition to object headers and length prefix all java objects also have to be aligned, usually to pointer sizes, possibly multiples thereof (there's an option to tune object alignment).

The need for alignment means there has to be padding end of each object - and thus the array - if the subsequent object wouldn't align properly. That concern will be most prominent with byte[] arrays.

the8472
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