为什么std :: make_unique <std::thread> （somefun（））在我没有调用release（）时会崩溃，而在调用时会崩溃吗？

Question

My code:

{
    t = std::make_unique<std::thread>(std::bind(&Widget::generateNum, this));
}

crash at ~thread() , error msg is ' r6010 abort() has been called '. If I haven't call t.release() before that destructing t , will cause the crash.

Answer 1

You must detach or join a thread before destroying it.

auto t = std::make_unique<std::thread>(std::bind(&Widget::generateNum, this));

// Either do this:
t->detach();
// or do this:
t->join();
// before *t gets destroyed.

The choice of whether to detach or join is up to you, but you must do one or the other. If a non-empty std::thread gets destroyed it will call std::terminate , ending your program.

See std::thread::~thread :

If *this has an associated thread ( joinable() == true ), std::terminate() is called.

The default std::terminate handler calls std::abort , hence the message that abort has been called.

Don't Call t.release()

It would be a mistake to call t.release() . This would leak the std::thread object.