Hans Passant provided this link which told me that when_any( ... ).then([]( <params> )
requires to be of the type std::pair<returnType, size_t index>
.
How does when_any work?
题
I have a std::vector< concurrency::task<void> >
with a list of tasks that may or may not load in order. I came across the concurrency::when_any
call, but don't know enough information about how to use it.
I came across this microsoft sample that uses the call (ScenarioInput1.xaml.cpp:100), but I don't understand the pair parameter, and how to adapt that to my return values.
Edit: What I'm trying to do:
struct TaskInfo {
unsigned int uniqueId;
concurrency::task<unsigned int> task;
};
typedef std::vector< TaskInfo > TaskList;
typedef std::vector< unsigned int > TaskReturns;
// Somewhere inside a class, tasks and finished are member variable.
TaskList tasks;
bool finished = false;
// Inside a member function of the same class as above
// Populate the tasks
concurrency::when_any( tasks.begin(), tasks.end() ).then([this](std::vector<unsigned int> ids){
for( std::vector<unsigned int>::iterator list=ids.begin; list != ids.end(); ++ids ) {
// a mutex lock would happen here
for( TaskList::iterator allTasks = tasks.begin(); allTasks != tasks.end(); ++allTasks ) {
if ( *list == allTasks->uniqueId ) { tasks.erase( allTasks ); break; }
}
// mutex unlock
if ( tasks.size() == 0 ) { finished = true; break }
}
}
// In another member function
if ( finished ) doOtherThings();
If what I'm trying to do is not as well thought out or not as efficient, please let me know.
解决方案 2
其他提示
It's the "choice" combination for tasks. You have a continuation (then), a join (when_all) and choice (when_any). The docs say ...
Creates a task that will complete successfully when any of the tasks supplied as arguments completes successfully.
See the task class and Asynchronous programming in C++ for more details. More discussion and examples.
Take for example:
task<string> tasks[] = {t1, t2, t3};
auto taskResult = when_all (begin(tasks), end(tasks))
.then([](std::vector<string> results) {
... is equivalent to ...
(t1 && t2 && t3).then
Whereas ...
task<string> tasks[] = {t1, t2, t3};
auto taskResult = when_any (begin(tasks), end(tasks))
.then([](string result) {
... is equivalent to ...
(t1 || t2 || t3).then