// This example demonstrates how to use cpp-taskflow to create
// dynamic workload during execution.
//
// We first create four tasks A, B, C, and D. During the execution
// of B, it uses flow builder to creates another three tasks
// B1, B2, and B3, and adds dependencies from B1 and B2 to B3.
//
// We use dispatch and get to wait until the graph finished.
// Do so is difference from "wait_for_all" which will clean up the
// finished graphs. After the graph finished, we dump the topology
// for inspection.
//
// Usage: ./subflow detach|join
//

#include <taskflow/taskflow.hpp>

const auto usage = "usage: ./subflow detach|join";

int main(int argc, char* argv[]) {

 if(argc != 2) {
 std::cerr << usage << std::endl;
 std::exit(EXIT_FAILURE);
 }

 std::string opt(argv[1]);

 if(opt != "detach" && opt != "join") {
 std::cerr << usage << std::endl;
 std::exit(EXIT_FAILURE);
 }

 auto detached = (opt == "detach") ? true : false;

 // Create a taskflow graph with three regular tasks and one subflow task.
 tf::Executor executor(4);
 tf::Taskflow taskflow("Dynamic Tasking Demo");

 // Task A
 auto A = taskflow.emplace([] () { std::cout << "TaskA\n"; });
 auto B = taskflow.emplace(
 // Task B
 [cap=std::vector<int>{1,2,3,4,5,6,7,8}, detached] (tf::Subflow& subflow) {
 std::cout << "TaskB is spawning B1, B2, and B3 ...\n";

 auto B1 = subflow.emplace([&]() { 
 printf(" Subtask B1: reduce sum = %d\n", 
 std::accumulate(cap.begin(), cap.end(), 0, std::plus<int>()));
 }).name("B1"); 
 
 auto B2 = subflow.emplace([&]() { 
 printf(" Subtask B2: reduce multiply = %d\n", 
 std::accumulate(cap.begin(), cap.end(), 1, std::multiplies<int>()));
 }).name("B2"); 
 
 auto B3 = subflow.emplace([&]() { 
 printf(" Subtask B3: reduce minus = %d\n", 
 std::accumulate(cap.begin(), cap.end(), 0, std::minus<int>()));
 }).name("B3"); 
 
 B1.precede(B3);
 B2.precede(B3);

 // detach or join the subflow (by default the subflow join at B)
 if(detached) subflow.detach();
 }
 );
 
 auto C = taskflow.emplace([] () { std::cout << "TaskC\n"; });
 auto D = taskflow.emplace([] () { std::cout << "TaskD\n"; });
 A.name("A");
 B.name("B");
 C.name("C");
 D.name("D");
 
 A.precede(B); // B runs after A 
 A.precede(C); // C runs after A 
 B.precede(D); // D runs after B 
 C.precede(D); // D runs after C 

 executor.run(taskflow).get(); // block until finished

 // examine the graph
 taskflow.dump(std::cout);

 return 0;
}