Very Computer

Hello.  I have a client/server application that uses sockets.  I've
run into a problem of the server read()'ing a socket and*.

After toying with the code a bit, I think I found out what's wrong:

Here's the problematic server code, in much shorter form:

for( i = 0; i < 3; i++)
  {
  read(...);
  do_Something();
  }

Here's the client code:
write(...);
write(...);
write(...);

The above code freezes the client and server, with the server*
on the second read.

Now if I change the client code to:

write(...);
sleep(2);
write(...);
sleep(2);
write(...);
sleep(2);

Everything is fine!  Why is that?  I'm just writing short strings
(less than 30-50 chars) with each write.  My understanding (perhaps
faulty) is that a write() by a client before the server is ready to
read() results in the information being held in a buffer?

If you can help me figure out what's going on, I'd greatly appreciate
it.  Thanks!

-Godfrey Degamo,

  You should also be prepared to receive packets which are broken up
into smaller parts--the client's stack may decide it only has room for
50 bytes at the moment when you told it you have 100 bytes to send...
your client then has to post the remaining 50 bytes with another call.

  Have fun!

-Bruce

----

Software Engineer, Project Manager
ROI Systems, Inc.                 801-255-1777 x106, or Fax: 255-9699

 Godfrey> Hello.  I have a client/server application that uses
 Godfrey> sockets.  I've run into a problem of the server read()'ing a
 Godfrey> socket and*.

 Godfrey> After toying with the code a bit, I think I found out what's
 Godfrey> wrong:

 Godfrey> Here's the problematic server code, in much shorter form:

 Godfrey> for( i = 0; i < 3; i++)
 Godfrey>   {
 Godfrey>   read(...);
 Godfrey>   do_Something();
 Godfrey>   }

 Godfrey> Here's the client code:
 Godfrey> write(...);
 Godfrey> write(...);
 Godfrey> write(...);

 Godfrey> The above code freezes the client and server, with the
 Godfrey> server* on the second read.

 Godfrey> Now if I change the client code to:

 Godfrey> write(...);
 Godfrey> sleep(2);
 Godfrey> write(...);
 Godfrey> sleep(2);
 Godfrey> write(...);
 Godfrey> sleep(2);

 Godfrey> Everything is fine!  Why is that?  I'm just writing short
 Godfrey> strings (less than 30-50 chars) with each write.  My
 Godfrey> understanding (perhaps faulty) is that a write() by a client
 Godfrey> before the server is ready to read() results in the
 Godfrey> information being held in a buffer?

You don't say, but I assume you're using stream sockets.

The problem is that the boundaries of the writes are lost; there is no
1-1 relationship between writes and reads. You will probably find that
all your data is being read in the first read() call.

ObFAQ:
You might want to take a look at the unix-socket-faq, regularly posted
here (and in news.answers & comp.answers), and also available at:
  http://www.veryComputer.com/
  http://www.veryComputer.com/~vic/sock-faq
  ftp://rtfm.mit.edu/pub/usenet/news.answers/unix-faq/socket

--

"Ceterum censeo Microsoftam delendam esse" - Alain Knaff in nanam

Quote:>My understanding (perhaps
>faulty) is that a write() by a client before the server is ready to
>read() results in the information being held in a buffer?

Boundaries between write() data are not known and not preserved.
The first read() will probably get the data from all 3 write()s,
leaving no data to read on the second/third read()s, hence you
block.

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