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

moongeegee <>
Sat, 28 Mar 2009 03:56:15 -0700 (PDT)
On Mar 27, 8:27 pm, Mark Space <> 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=


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


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=


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=


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

Thank you very much for your help.
Kee Kee

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