# Communication Between Graphs

All graph nodes require a reference to a graph object as one of the arguments to their constructor. It is only safe to construct edges between nodes that are part of the same graph. An edge expresses the topology of your graph to the runtime library. Connecting two nodes in different graphs can make it difficult to reason about whole graph operations, such as calls to graph::wait_for_all and exception handling. To optimize performance, the library may make calls to a node's predecessor or successor at times that are unexpected by the user.

If two graphs must communicate, do NOT create an edge between them, but instead use explicit calls to try_put. This will prevent the runtime library from making any assumptions about the relationship of the two nodes, and therefore make it easier to reason about events that cross the graph boundaries. However, it may still be difficult to reason about whole graph operations. For example, consider the graphs below:

```    graph g;
function_node< int, int > n1( g, 1, [](int i) -> int {
cout << "n1\n";
spin_for(i);
return i;
} );
function_node< int, int > n2( g, 1, [](int i) -> int {
cout << "n2\n";
spin_for(i);
return i;
} );
make_edge( n1, n2 );

graph g2;
function_node< int, int > m1( g2, 1, [](int i) -> int {
cout << "m1\n";
spin_for(i);
return i;
} );
function_node< int, int > m2( g2, 1, [&](int i) -> int {
cout << "m2\n";
spin_for(i);
n1.try_put(i);
return i;
} );
make_edge( m1, m2 );

m1.try_put( 1 );

// The following call returns immediately:
g.wait_for_all();
// The following call returns after m1 & m2
g2.wait_for_all();

// we reach here before n1 & n2 are finished
// even though wait_for_all was called on both graphs
```

In the example above, m1.try_put(1) sends a message to node m1, which runs its body and then sends a message to node m2. Next, node m2 runs its body and sends a message to n1 using an explicit try_put. In turn, n1 runs its body and sends a message to n2. The runtime library does not consider m2 to be a predecessor of n1 since no edge exists.

If you want to wait until all of the tasks spawned by these graphs are done, you need to call the function wait_for_all on both graphs. However, because there is cross-graph communication, the order of the calls is important. In the (incorrect) code segment above, the first call to g.wait_for_all() returns immediately because there are no tasks yet active in g; the only tasks that have been spawned by then belong to g2. The call to g2.wait_for_all returns after both m1 and m2 are done, since they belong to g2; the call does not however wait for n1 and n2, since they belong to g. The end of this code segment is therefore reached before n1 and n2 are done.

If the calls to wait_for_all are swapped, the code works as expected:

```    g2.wait_for_all();
g.wait_for_all();