为什么在将move构造函数传递给async时，会从默认构造函数中调用它？

#include <iostream>
#include <future>

class MyClass {
public:
    MyClass() { std::cout << "Default constructor" << std::endl; }
    MyClass(MyClass&& other) noexcept { std::cout << "Move constructor" << std::endl; }
};

int main() {
    MyClass obj;
    auto future = std::async(std::launch::async, [&obj]() {
        MyClass movedObj(std::move(obj));
        // 使用movedObj进行操作
    });
    future.get();
    return 0;
}

#include <iostream>
#include <future>

class MyClass {
public:
    MyClass() { std::cout << "Default constructor" << std::endl; }
    MyClass(MyClass&& other) noexcept { std::cout << "Move constructor" << std::endl; }
};

int main() {
    MyClass obj;
    std::packaged_task<void()> task([&obj]() {
        MyClass movedObj(std::move(obj));
        // 使用movedObj进行操作
    });
    auto future = task.get_future();
    std::thread(std::move(task)).detach();
    future.get();
    return 0;
}