My understanding is that when an asynchronous operation throws an exception, it will be propagated back to a thread that calls std::future::get()
. However, when such a thread calls std::future::wait()
, the exception is not immediately propagated - it'll be thrown upon a subsequent call to std::future::get()
.
However, In such a scenario, what is supposed to happen to such an exception if the future object goes out of scope after a call to std::future::wait()
, but prior to a call to std::future::get()
?
For those interested, here is a simple example. In this case, the exception is silently handled by the thread/future package:
#include "stdafx.h"
#include <thread>
#include <future>
#include <iostream>
int32_t DoWork( int32_t i )
{
std::cout << "i == " << i << std::endl;
throw std::runtime_error( "DoWork test exception" );
return 0;
}
int _tmain(int argc, _TCHAR* argv[])
{
auto f = std::async( DoWork, 5 );
try
{
//f.get(); // 1 - Exception does propagate.
f.wait(); // 2 - Exception does NOT propagate.
}
catch( std::exception& e )
{
std::cout << e.what() << std::endl;
return -1;
}
return 0;
}
vs2012\vc11\crt\future.cpp
there is an error with the
this code generated an invalid acceso to "_Future_messages" because _Mycode.value() return 4.
// code example
It is ignored and discarded, just like if you
wait()
for a value but neverget()
it.wait()
simply says "block until the future is ready", be that ready with a value or exception. It's up to the caller to actuallyget()
the value (or exception). Usually you'll just useget()
, which waits anyway.