Issue invoking member function asynchronously when class has a std::mutex member

Question

Hi

The code below compiles and runs successfully and Foo's compute method can be invoked both synchronously as well as asynchronously using std::async.

#include <iostream>
#include <future>

struct Foo
{
    int compute()
    {
        return 123;
    }

    // === UNCOMMENT THE LINE BELOW TO SHOW THE ERROR ===
    //std::mutex the_mutex;
};

int main()
{
    Foo foo;
    std::cout << "result: " << foo.compute() << std::endl;

    std::future<int> f = std::async(std::launch::async, &Foo::compute, foo);
    std::cout << "result: " << f.get() << std::endl;

    return 0;
}

However, if you uncomment the std::mutex member, compilation fails with the following error:

Error C2661 'std::tuple<int (__thiscall Foo:: )(void),Foo>::tuple': no overloaded function takes 2 arguments*

Moreover, the error disappears if you comment out the code that invokes the method asynchronously, I'm using VS2019 v16.7.4. Note that this is not an issue with std::mutex per se. I can see the same behavior if I replaced it with a boost::asio::io_context member variable.

I haven't done C++ programming for a very long time. Assuming this is not a compiler bug, I would like to understand the reason and find a workaround.

Thanks

Answer 1

Hi,

std::mutex is non-copyable. Therefore, the Foo class that contains it becomes non-copyable. You are passing a temporary Foo object to std::future, and you probably want to use the same object in all your threads. I suggest you could use std::ref to pass a reference of an existing object.

Here is the code:

#include <iostream>  
#include <future>  
  
struct Foo  
{  
	int compute()  
	{  
	return 123;  
	}  
  
	// === UNCOMMENT THE LINE BELOW TO SHOW THE ERROR ===  
	std::mutex the_mutex;  
};  
  
int main()  
{  
	Foo foo;  
	std::cout << "result: " << foo.compute() << std::endl;  
  
	std::future<int> f = std::async(std::launch::async, &Foo::compute, std::ref(foo));  
	std::cout << "result: " << f.get() << std::endl;  
  
	return 0;  
}  

Best Regards,

Jeanine Zhang

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