Exception propagation and std::future

Question

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;
}

Answer 1

It is ignored and discarded, just like if you wait() for a value but never get() 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 actually get() the value (or exception). Usually you'll just use get(), which waits anyway.