Input and Output - Java 8 Recipes - page 297

After the server acquires a socket channel, it creates an inputStream by calling
Channels.newInputStream(socket) and then wrapping that input stream
with an ObjectInputStream . The server then proceeds to loop and read each ob-
ject coming from the ObjectInputStream . If the object received's toString()
method equals EOF , the server stops looping and the connection is closed.
Note Using an ObjectOutputStream and ObjectInputStream to send and
receive a lot of objects can lead to memory leaks. ObjectOutputStream keeps a
copy of the sent object for efficiency. If you were to send the same object again, Ob-
jectOutputStream and ObjectInputStream will not send the same object
again, but instead send a previously sent Object ID. This behavior or just sending the
Object ID instead of the whole object raises two issues.
The first issue is that objects that are changed in-place (mutable) will not get the change
reflected in the receiving client when sent through the wire. The reason is that because
the object was sent once, the ObjectOutputStream believes that the object is
already transmitted and will only send the ID, negating any changes to the object that
have happened since it was sent. To avoid this, don't make changes to objects that were
sent down the wire. This rule also applies to sub-objects from the object graph.
The second issue is that because ObjectOutputStream maintains a list of sent ob-
jects and their Object IDs, if you send a lot of objects the dictionary of sent objects to
keys grows indefinitely, causing memory starvation on a long-running program. To alle-
viate this issue, you can call ObjectOutputStream.reset() , which will clear
the dictionary of sent objects. Alternatively, you can invoke ObjectOut-
putStream.writeUnshared() to not cache the object in the ObjectOut-
putStream dictionary.
8-5. Obtaining the Java Execution Path
Problem
You want to get the path where the Java program is running.
Solution