On 21 Aug 2001 13:47:50 +0200, Dalibor Topic wrote:
> Am Montag, 20. August 2001 17:51 schrieb Tom Tromey:
> I completely understand that. It's a cycle-burner. It's not elegant. It was
> more of a proof-of-concept that you could fix the bug, without the need for
> internal threads etc. I have a more elegant solution using wait and notify
> which is quite close to the solution proposed by John. I wrote a simple test
> class WaitingPushbackInputsStream which extends PushbackInputStream. It has
> the following methods:
>
> public class WaitingPushbackInputStream extends PushbackInputStream {
> [...]
> public int read() throws IOException {
> synchronized(this) {
> synchronized(in) {
> while(0 == available()) {
> try {
> wait();
> }
> catch (InterruptedException e) {
> // ignore
> }
> }
>
> // something is available for reading,
> // and we've got both locks. this read
> // should not block at all.
> return super.read();
> }
> }
> }
>
> public void unread(int oneByte) throws IOException {
> synchronized(this) {
> notifyAll();
> super.unread(oneByte);
> }
> }
> }
>
> this solution avoids polling, and only reads when there is something
> available for reading. It relies on being notified that there is something
> available. From what I've seen in my test, it doesn't burn cycles while it's
> waiting.
>
> I used notifyAll(), but in this case where you have a single byte to be
> (un)read, notify() should be just fine. I think that you need notifyAll() if
> you push back several bytes, since the notified thread might not want to read
> all bytes you pushed back. Then you'd be left with some threads waiting
> although there are bytes available in the pushback buffer.
>
> The call to available() should take care of someone closing the
> PushbackInputStream or the sub-stream while a thread is waiting to read.
>
That's some nice code. But how could super.unread() ever get called if
read() is in its loop? They're both synchronized on this.
--John
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