Visible to Intel only — GUID: GUID-8DC6237A-BE3C-4F6A-BA98-C3EAEF9A13F5
Package Contents
Parallelizing Simple Loops
Parallelizing Complex Loops
Parallelizing Data Flow and Dependence Graphs
Work Isolation
Exceptions and Cancellation
Floating-point Settings
Containers
Mutual Exclusion
Timing
Memory Allocation
The Task Scheduler
Design Patterns
Migrating from Threading Building Blocks (TBB)
Constrained APIs
Appendix A Costs of Time Slicing
Appendix B Mixing With Other Threading Packages
References
parallel_for_each Body semantics and requirements
parallel_sort ranges interface extension
TBB_malloc_replacement_log Function
Type-specified message keys for join_node
Scalable Memory Pools
Helper Functions for Expressing Graphs
concurrent_lru_cache
task_group extensions
The customizing mutex type for concurrent_hash_map
Visible to Intel only — GUID: GUID-8DC6237A-BE3C-4F6A-BA98-C3EAEF9A13F5
Destroying Graphs That Run Outside the Main Thread
Make sure to enqueue a task to wait for and destroy graphs that run outside the main thread.
You may not always want to block the main application thread by calling wait_for_all(). However, it is safest to call wait_for_all on a graph before destroying it. A common solution is to enqueue a task to build and wait for the graph to complete. For example, assume you really do not want to call a wait_for_all in the example from Always Use wait_for_all(), Instead you can enqueue a task that creates the graph and waits for it:
class background_task {
public:
void operator()() {
graph g;
function_node< int, int > f( g, 1, []( int i ) -> int {
return spin_for(i);
} );
f.try_put(1);
g.wait_for_all();
}
};
void no_wait_for_all_enqueue() {
task_arena a;
a.enqueue(background_task());
// do other things without waiting…
}
In the code snippet above, the enqueued task executes at some point, but it’s not clear when. If you need to use the results of the enqueued task, or even ensure that it completes before the program ends, you will need to use some mechanism to signal from the enqueued task that the graph is complete.