C++ 如何在rxcpp中处理请求/响应流
我需要在rxcpp中实现一个摄像头采样系统。。我想象的方式是将requestStream作为param传递,然后接收responseStream 每次调用requestSample时,都会创建一个新的摄像头会话,当为requestStream调用on_complete()时,摄像头会话将终止C++ 如何在rxcpp中处理请求/响应流,c++,c++11,reactive-programming,rxcpp,C++,C++11,Reactive Programming,Rxcpp,我需要在rxcpp中实现一个摄像头采样系统。。我想象的方式是将requestStream作为param传递,然后接收responseStream 每次调用requestSample时,都会创建一个新的摄像头会话,当为requestStream调用on_complete()时,摄像头会话将终止 observable<ImageSample> requestSampleStream(observable<ImageRequest> requestStream$) { au
observable<ImageSample> requestSampleStream(observable<ImageRequest> requestStream$) {
auto response$ = rxcpp::observable<>::create<ImageSample>(
[&](rxcpp::subscriber<ImageSample> s){
auto request_next = [&](ImageRequest imageRequest) {
cout << "image request next" << endl;
SampleImage vsi;
s.on_next(vsi);
};
auto request_completed = [&] {
cout << "no more samples needed.. close camera" << endl;
s.on_completed();
};
auto request_error = [&](std::exception_ptr e) {
try { rethrow_exception(e); }
catch (const exception &ex) {
cerr << "error happened on request stream.. close the camera and send error on return stream" << endl << ex.what() << endl;
}
s.on_error(e);
};
requestStream$.subscribe(request_next,
request_error,
request_completed
);
});
return response$;
}
observable requestSampleStream(observable requestStream$){
自动响应$=rxcpp::可观察::创建(
[&](rxcpp::订户s){
自动请求\u next=[&](ImageRequest ImageRequest){
couts
是一个本地堆栈变量,但lambda正在捕获对s
的引用。调用lambda时,此引用无效。将[&]
更改为[=]
,这样应该可以工作
另一个选项是使用现有的map
algo
observable<ImageSample> requestSampleStream(observable<ImageRequest> requestStream$) {
return requestStream$ |
map([](ImageRequest imageRequest){
cout << "image request next" << endl;
SampleImage vsi;
s.on_next(vsi);
})
// optional
| tap([&](std::exception_ptr e) {
try { rethrow_exception(e); }
catch (const exception &ex) {
cerr << "error happened on request stream.. close the camera and send error on return stream" << endl << ex.what() << endl;
}
},
[](){
cout << "no more samples needed.. close camera" << endl;
})
;
}
observable requestSampleStream(observable requestStream$){
返回请求流$|
映射([](ImageRequest ImageRequest){
cout Spot on..什么是rxcpp-ish更有效的方法?通过map+tap?是的,使用现有的map算法比使用create来复制map要好。我唯一的问题是map的结果是异步的,所以map无法同步返回..有解决方法吗?通常使用flat_map来获得异步结果。'flat_map'只是'map()| merge()'其中映射返回一个“可观察的”