有没有一种创建线程的方法，但是可以使用C ++ 11延迟执行而不使用operator =

Question

two threads are created if i use the following:

void func()
{}

std::thread td1;//thread  object created.
...
...
//defer the running of td1 until now
td1 = std::thread(func);//temp thread object(rvalue) created

Is there a way to achieve the defer execution but without creating two threads?

Answer 1

I'm not sure what the point is, but...

No operator= , though two threads are constructed:

std::thread td1;
...
std::thread(func).swap(td1);

Or no operator= and only one thread constructed:

std::promise<void> p;
std::thread td1([&]() {
   p.get_future().wait();
   func();
});
...
p.set_value();

td1.join(); // Can't let p go out of scope pending the thread.

Maybe what you really want is:

std::unique_ptr<std::thread> td1;
...
td1.reset(new std::thread(func));

Answer 2

std::thread td1;

That creates a thread object, but it does not create a thread. So:

std::thread td1;
td1 = std::thread(func);

is perfectly fine code even if you are concerned about performance.