Re: Is a byte data type really a 32-bit int in the JVM?

From:

=?ISO-8859-1?Q?Arne_Vajh=F8j?= <arne@vajhoej.dk>

Newsgroups:

comp.lang.java.programmer

Date:

Sun, 03 Feb 2008 15:57:06 -0500

Message-ID:

<47a62a9d$0$90265$14726298@news.sunsite.dk>

Mike Schilling wrote:

Digital Puer wrote:

Is a byte data type really a 32-bit int in the JVM? More
specifically, if I have an an array of N byte types, are N
32-bit ints actually allocated underneath? I am writing
a memory-sensitive application and would appreciate
some insight.

The real answer is "That's up to the JVM implementer; within the
language, there's no way to tell." Note that is a very different
answer than you'd get for C or C++, where there are lots of ways to
tell, sizeof(char) being the simplest one.

JVM implementers not being idiots, and arrays of bytes being used all
over the place in the system classes, I very much doubt that any JVM
implementation makes them four times as big as they have to be. It is
(IIRC) a common implementation that bytes used as local variables take
up a full 32 bits, but in normal cases that's a small enough fraction
of the total size of a new stack frame to be down in the noise.

A more interesting question is whether each entry in an array of
boolean takes up a full byte, when in principle only a single bit is
needed. An implementer needs to weigh the added cost of accessing a
single bit against the savings in space. I don't know what the usual
result is.

public class Sizeof {
     private final static int N = 10000000;
     public static long mem() {
         System.gc();
         Runtime rt = Runtime.getRuntime();
         return rt.totalMemory() - rt.freeMemory();
     }
     public static void main(String[] args) {
         long m1 = mem();
         byte[] ba = new byte[N];
         long m2 = mem();
         System.out.println("sizeof byte = " + (m2 - m1)*1.0/N);
     }
}

will give a hint !

(it indicates 1 for the Java version I am using)

Arne

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