Calling std::async twice without storing the returned std::future

Question

According to the C++17 standard, what is the output of this program?

#include <iostream>
#include <string>
#include <future>

int main() {
  std::string x = "x";

  std::async(std::launch::async, [&x]() {
    x = "y";
  });
  std::async(std::launch::async, [&x]() {
    x = "z";
  });

  std::cout << x;
}

The program is guaranteed to output: z?

Answer 1

C++ reference explicitly mentions the behavior of this code:

If the std::future obtained from std::async is not moved from or bound to a reference, the destructor of the std::future will block at the end of the full expression until the asynchronous operation completes, essentially making code such as the following synchronous:

std::async(std::launch::async, []{ f(); }); // temporary's dtor waits for f()
std::async(std::launch::async, []{ g(); }); // does not start until f() completes

So your code is guaranteed to print z - there are no data races.