2.5.2 多线程get一个future 在如下代码中,我写了一个void future_get_func(std::future<int>& fu)的函数,尝试开一个线程来get,然后主执行流又get一次,看看会发生什么。 #include <iostream> #include <future> #include <functional> #include <thread> #include <type_traits> // result_of #include <s...
voidwait()const; 当共享状态值是不可以用时,调用wait接口可以一直阻塞,直到共享状态变为"就绪"时,就变为可以用了。 get操作 get是获取共享状态的结果它有以下三种形式: //仅为泛型 future 模板的成员Tget();//(仅为 future<T&> 模板特化的成员)T&get();//仅为 future<void> 模板特化的成员voidget();...
std::future<void> fut = std::async(throw_exception); try { fut.get(); } catch (const std::exception& e) { std::cout << "Caught exception: " << e.what() << std::endl; } return 0; }输出结果:Caught exception: Exception thrown<future> 库为C++ 程序员提供了一种简单而强大的异步...
1#include <iostream>//std::cout2#include <functional>//std::ref3#include <thread>//std::thread4#include <future>//std::promise, std::future5voidprint_int(std::future<int>&fut) {6intx = fut.get();//获取共享状态的值.7std::cout <<"value:"<< x <<'\n';//打印 value: 10.8}...
void SkiaGLRenderEngine::drawLayersInternal( const std::shared_ptr<std::promise<RenderEngineResult>>&& resultPromise, const DisplaySettings& display, const std::vector<LayerSettings>& layers, const std::shared_ptr<ExternalTexture>& buffer, const bool /*useFramebufferCache*/, base::unique_fd&& ...
auto f2 = std::async([=] {doSomething(c); }); 1. 2. 3. 4. 5. 6. 7. 8. 例如下面是对函数的传值调用: void doSomething(char c); //传值调用 auto f1 = std::async(doSomething, '.'); //传值调用 char c = '+';
template <class T> future; template <class R&> future<R&>; // specialization : T is a reference type (R&) template <> future<void>; // specialization : T is void Future Afutureis an object that can retrieve a value from some provider object or function, properly synchronizing this ...
#include<iostream>#include<thread>#include<future>voidmodifyMessage(std::promise<std::string>&&proms,std::string msg){std::string metaMsg=msg+" has been modified";proms.set_value(metaMsg);}intmain(){std::string msg_str="My Message";//创建promise对象std::promise<std::string>proms;//创建...
void main() { new Future.delayed(new Duration(seconds: 3), () { return 1; }).timeout(new Duration(seconds: 2)).then(print).catchError(print); } 运行上述代码会看到: TimeoutException after 0:00:02.000000: Future not completed。
CompletableFuture<Void> thenAcceptAsync(Consumer<? super T> action) Returns a new CompletionStage that, when this stage completes normally, is executed using this stage's default asynchronous execution facility, with this stage's result as the argument to the supplied action....