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Q: How big is an Object? Why is there no sizeof?

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Q: How big is an Object? Why is there no sizeof?

Ansswer: C has a sizeof operator, and it needs to have one, because the user has to manage calls to malloc, and because the size of primitive types (like long) is not standardized. Java doesn't need a sizeof, but it would still have been a convenient aid. Since it's not there, you can do this:

static Runtime runtime = Runtime.getRuntime();
...
long start, end;
Object obj;
runtime.gc();
start = runtime.freememory();
obj = new Object(); // Or whatever you want to look at
end =  runtime.freememory();
System.out.println("That took " + (start-end) + " 
bytes.");

This method is not foolproof, because a garbage collection could occur in the middle of the code you are instrumenting, throwing off the byte count. Also, if you are using a just-in-time compiler, some bytes may come from generating code.

You might be surprised to find that an Object takes 16 bytes, or 4 words, in the Sun JDK VM. This breaks down as follows: There is a two-word header, where one word is a pointer to the object's class, and the other points to the instance variables. Even though Object has no instance variables, Java still allocates one word for the variables. Finally, there is a "handle", which is another pointer to the two-word header. Sun says that this extra level of indirection makes garbage collection simpler. (There have been high performance Lisp and Smalltalk garbage collectors that do not use the extra level for at least 15 years. I have heard but have not confirmed that the Microsoft JVM does not have the extra level of indirection.)

An empty new String() takes 40 bytes, or 10 words: 3 words of pointer overhead, 3 words for the instance variables (the start index, end index, and character array), and 4 words for the empty char array. Creating a substring of an existing string takes "only" 6 words, because the char array is shared. Putting an Integer key and Integer value into a Hashtable takes 64 bytes (in addition to the four bytes that were pre-allocated in the Hashtable array): I'll let you work out why.


This tip is reprinted on JavaFAQ.nu by by courtesy of Peter Norvig I am thankful for his important contributions to my site - 21 Infrequently Answered Java Questions. Alexandre Patchine


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