On 7/20/06, Geir Magnusson Jr <[EMAIL PROTECTED]> wrote:
Mikhail Fursov wrote: > Add this line to the test to clarify the situation: > System.out.println(ss.getInetAddress()); > > I think that for PC with multiple network adapters serversocket IP must be > specified manually. I don't think so. "0.0.0.0" should mean all of them, right?
Yes. a socket listens on "0.0.0.0" means on it listens on all of network interfaces on this machine. It's often used for binding function. But the behaviour of connecting to "0.0.0.0" is unspeicified. At least, windows treats it as invalid target address, while linux allows it. geir
On 7/20/06, Victor A Martynov <[EMAIL PROTECTED]> wrote:Hello, I'm having trouble with an application. This problem concerns creating java.net.Socket on host "0.0.0.0". The test below illustrates this problem: import java.net.*; public class Test { public static void main(String args[]) throws Throwable { ServerSocket ss = new ServerSocket(1099); Socket s = new Socket("0.0.0.0", 1099); } } The output on RI and on Harmony differs: c:\test>java Test c:\test>C:/Harmony/bin/ij Test java.net.ConnectException: 0.0.0.0/0.0.0.0:1099 - The address is not available at org.apache.harmony.luni.net.PlainSocketImpl.connect( PlainSocketImpl.java:238) at org.apache.harmony.luni.net.PlainSocketImpl.connect( PlainSocketImpl.java:200) at java.net.Socket.startupSocket(Socket.java:715) at java.net.Socket.<init>(Socket.java:139) at Test.main(Test.java:7) I wonder whether it is a bug in RI or a bug in LUNI of Harmony? ===== Warm Regards, Victor Martynov, Intel Middleware Product Division
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