[mpich-discuss] //Ssend(...) safe?

[Sören Freudiger]

Hi @All
A hopefully very easy question. I just want to ensure that there's no misunderstanding:

/////////////////////////////////////////////////////////////////////////////////
if(rank==0)
{
   MPI::Request request = comm.Irecv(recvData, recvData.size, MPI__INT, 1, 100)
   /* do something */
   request.Wait();
   request = comm.Irecv(...);
   /* do something */
   request.Wait();
}
else
{
   /* do something */
   comm.Ssend(sendData, sendData.size, MPI::INT, 0, 100);
   sendData[0] = 100;
   comm.Ssend(sendData, sendData.size, MPI::INT, 0, 100);         
}
 
comm.Barrier();
/////////////////////////////////////////////////////////////////////////////////

I read somewhere:

" When a synchronous mode send operation is completed, the sending process may assume the destination process has begun receiving the message. The destination process need not be done receiving the message, but it must have begun receiving the message. "

Does that mean that under certain circumstances sendData[0] = 100; changes the data that is send at my first comm.Ssend(..) call? 
I think it's like that:
On the send side the send operation is completed as soon as the other side has started the receiving process. Nevertheless if all Data has been received on the other side there's no risk to write into the send buffer at the sending side again. That means: no chance that sendData[0] = 100; at first Ssend-operation. Am I correct?

Best,
DarkAnn

PS: I'm using Ssend, because possibly my send-call happens before my Irecv has been called. That means Rsend results in undefined behaivoir and that means standard Send shouldn't been used because reday send could be implemented for that one.