Issue1348645

With Jython 2.1, I tried to run the send() method from
socket.py and got the error :    
   
   File "C:\jython\Lib\socket.py", line 304, in send
TypeError: sendto() takes at least 3 arguments (2 given)

Here is the method in socket.py with the problem shown
in the stacktrace:

   def send(self, data):
	assert self.addr
	return self.sendto(self.addr)      # line 304


The definition of self.sendto() is:

     def sendto(self, data, addr):
	n = len(data)
	if not self.sock:
	    self.sock = java.net.DatagramSocket()
	host, port = addr
	bytes = jarray.array(map(ord, data), 'b')
	a = java.net.InetAddress.getByName(host)
	packet = java.net.DatagramPacket(bytes, n, a, port)
	self.sock.send(packet)
	return n


It looks like there is a missing argument in the call
to self.sendto() that is in send().  If I edit send():

	return self.sendto(data, self.addr)   # line 304
	
it seems to fix the problem.

Logged In: YES 
user_id=193969

Alan Kennedy is working on some updates to sockets from
jython -- so hopefully this will get cleaned up in the 2.2
series.

This was indeed a bug; the send function could never have worked as it was.

This is already fixed in 2.2b2.

As for jython 2.1, the simple fix suggested above will work, i.e. change the send function to be defined as

def send(self, data):
    assert self.addr
    return self.sendto(data, self.addr)