Flow Graph Basics: Reservation#
oneAPI Threading Building Blocks (oneTBB) flow graph
join_node
has four possible policies: queueing
, reserving
,
key_matching
and tag_matching
. join_nodes
need messages at
every input before they can create an output message. The reserving
join_node
does not have internal buffering, and it does not pull
messages from its inputs until it has a message at each input. To create
an output message it temporarily reserves a message at each input port,
and only if all input ports succeed reserving messages will an output
message be created. If any input port fails to reserve a message, no
message will be pulled by the join_node
.
To support the reserving join_node
some nodes support
reservation of their outputs. The way reservation works is:
When a node connected to a reserving
join_node
in push state tries to push a message, thejoin_node
always rejects the push and the edge connecting the nodes is switched to pull mode.The reserving input port calls
try_reserve
on each edge in pull state. This may fail; if so, the reserving input port switches that edge to push state, and tries to reserve the next node connected by an edge in pull state. While the input port’s predecessor is in reserved state, no other node can retrieve the reserved value.If each input port successfully reserves an edge in pull state, the reserving
join_node
will create a message using the reserved messages and try to push the resulting message to any nodes connected to it.If the message is successfully pushed to a successor, the predecessors that were reserved are signaled that the messages were used (by calling
try_consume()
.) Those messages will be discarded by the predecessor nodes, because they have been successfully pushed.If the message was not successfully pushed to any successor, the predecessors that were reserved are signaled that the messages were not used (by calling
try_release()
.) At this point, the messages may be pushed to or pulled by other nodes.
Because the reserving join_node
will only attempt to push when each
input port has at least one edge in a pull state, and will only attempt
to create and push a message if all input ports succeed reserving
messages, at least one of the predecessors to each of the reserving
join_node
input ports must be reservable.
The following example demonstrates a reserving join_node
’s behavior.
buffer_nodes
buffer their output, so they accept a switch of their
output edge from push to pull mode. broadcast_nodes
do not buffer
messages and do not support try_get()
or try_reserve()
.
void run_example2() { // example for Flow_Graph_Reservation.xml
graph g;
broadcast_node<int> bn(g);
buffer_node<int> buf1(g);
buffer_node<int> buf2(g);
typedef join_node<tuple<int,int>, reserving> join_type;
join_type jn(g);
buffer_node<join_type::output_type> buf_out(g);
join_type::output_type tuple_out;
int icnt;
// join_node predecessors are both reservable buffer_nodes
make_edge(buf1,input_port<0>(jn));
make_edge(bn,input_port<0>(jn)); // attach a broadcast_node
make_edge(buf2,input_port<1>(jn));
make_edge(jn, buf_out);
bn.try_put(2);
buf1.try_put(3);
buf2.try_put(4);
buf2.try_put(7);
g.wait_for_all();
while (buf_out.try_get(tuple_out)) {
printf("join_node output == (%d,%d)\n",get<0>(tuple_out), get<1>(tuple_out) );
}
if(buf1.try_get(icnt)) printf("buf1 had %d\n", icnt);
else printf("buf1 was empty\n");
if(buf2.try_get(icnt)) printf("buf2 had %d\n", icnt);
else printf("buf2 was empty\n");
}
In the example above, port 0 of the reserving join_node
jn
has
two predecessors: a buffer_node
buf1
and a broadcast_node
bn
. Port 1 of the join_node
has one predecessor, buffer_node
buf2
.
We will discuss one possible execution sequence (the scheduling of tasks may differ slightly, but the end result will be the same.)
bn.try_put(2);
bn
attempts to forward 2 to jn
. jn
does not accept the value
and the arc from bn
to jn
reverses. Because neither bn nor jn
buffer messages, the message is dropped. Because not all the inputs to
jn
have available predecessors, jn
does nothing further.
Caution
Any node which does not support reservation will not work correctly
when attached to a reserving join_node
. This program demonstrates
why this occurs; connecting non-reserving nodes to nodes requiring
support for reservation is not recommended practice.
buf1.try_put(3);
buf1
attempts to forward 3 to jn
. jn
does not accept the
value and the arc from buf1
to jn
reverses. Because not all the
inputs to jn
have available predecessors, jn
does nothing
further.
buf2.try_put(4);
buf2
attempts to forward 4 to jn
. jn
does not accept the
value and the arc from buf2
to jn
reverses. Now both inputs of
jn
have predecessors, a task to build and forward a message from
jn
will be spawned. We assume that task is not yet executing.
buf2.try_put(7);
buf2
has no successor (because the arc to jn
is reversed,) so it
stores the value 7.
Now the task spawned to run jn
runs.
jn
tries to reservebn
, which fails. The arc tobn
switches back to the forward direction.jn
tries to reservebuf1
, which succeeds (reserved nodes are colored grey.)jn
receives the value 3 frombuf1
, but it remains inbuf1
(in case the attempt to forward a message fromjn
fails.)jn
tries to reservebuf2
, which succeeds.jn
receives the value 4 frombuf2
, but it remains inbuf2
.jn
constructs the output messagetuple<3,4>
.
Now jn
pushes its message to buf_out
, which accepts it. Because
the push succeeded, jn
signals buf1
and buf2
that the
reserved values were used, and the buffers discard those values. Now
jn
attempts to reserve again.
No attempt to pull from
bn
is made, because the edge frombn
tojn
is in push state.jn
tries to reservebuf1
, which fails. The arc tobuf1
switches back to the forward direction.jn
does not try any further actions.
No further activity occurs in the graph, and the wait_for_all()
will
complete. The output of this code is
join_node output == (3,4)
buf1 was empty
buf2 had 7