What happens if I declare a Queue of messages instead of a Queue of message
pointers ?
I am trying to find out the potential problems of declaring
QueueC(message_t, 8) instead of QueueC(message_t*, 8). Then, when I get a
packet from the lower layer, I instantiate a new message_t variable, copy
the contents of the received message there, queue it and return the buffer
to radio Layer.
event message_t* Receive.receive(message_t* bufPtr, void* payload, uint8_t
len) {
message_t tempMsg;
memcpy(&tempMsg,bufPtr,sizeof(message_t);
SendQueue.enqueue(tempMsg);
return bufPtr;
}
On 4/13/07, Philip Levis <[EMAIL PROTECTED]> wrote:
On Apr 12, 2007, at 7:25 PM, Mehedi Bakht wrote:
> Hi,
>
> I am a bit confused about why and how the interface Pool should be
> used when using the interface Queue (TinyOS 2.x). Any comment/
> explanation will be really appreciated.
>
The two are completely separate. It just happens to be that some use
cases require that you use both of them.
A Queue gives you a fixed-length queue of type <t>. For example, a
transmission queue is a Queue<message_t*>, that is, a queue of
pointers to message buffers. You would instantiate it an 8-deep send
queue as a QueueC(message_t*, 8).
A Pool is a fixed-size memory pool of type <t>. For example, a pool
of message buffers is a Pool<message_t>, and you instantiate it with
PoolC(message_t, 8).
It happens to be that packet forwarding layers need both of them:
when you receive a packet, you need to allocate one to give back to
the radio layer, and you also need to put the received packet into
the send queue.
Phil
--
Mehedi Bakht
Graduate Student
Department of Computer Science
University of Illinois at Urbana-Champaign
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