Re: byte[] out = new byte[MAX_NUM];

From:
moongeegee <moongeegee@gmail.com>
Newsgroups:
comp.lang.java.help
Date:
Sat, 28 Mar 2009 03:56:15 -0700 (PDT)
Message-ID:
<2253aa4e-137a-4337-b780-ba6400c4b602@o11g2000yql.googlegroups.com>
On Mar 27, 8:27 pm, Mark Space <marksp...@sbc.global.net> wrote:

moongeegee wrote:

I have java code as byte[] buf = new byte[MAX_NUM];

Should I free the byte[] after that? If I need to free the byte[], how
to free it.

I know C++, I need to do "delete [] byte", how about java?


It's normally not necessary. The array you allocated with "new" will b=

e

freed (or more properly garbage collected) when all of its references
are no longer reachable. But there are some special applications wher=

e

you need to "free" some object to prevent it being held when no longer
needed. So:

     buf = null;

will replace the reference "buf" with null, and the garbage collector
will clean up your object in a bit. That's it, you don't need to call
any special method.

Why would you do this? The common example is a stack. Imagine a sta=

ck

implemented as:

public class MyStack { // untested code...
     Object[] stack = new Object[20];
     int top;
     public void push( Object o ) {
         stack[top++] = o;
     }
     public Object pop() {
         return stack[--top];
     }

}

If you look carefully at the pop() method, you will notice that the
array "stack" still holds a reference to whatever object was just
returned. The reference was copied, and stack still has a copy. Thi=

s

counts as a memory leak in Java, because you're holding objects when you
really don't need to. The usual fix goes something like this:

     public Object pop() { // also untested
         Object o = stack[--top];
         stack[top] = null;
         return o;
     }

And now at least we aren't holding a reference when we don't need it
anymore.


Thank you very much for your help.
Kee Kee

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