Re: [Boost-users] [Thread] Can't compile package_task code with 1.54

Le 01/06/13 14:38, Vicente J. Botet Escriba a écrit :

Le 01/06/13 13:53, Vicente J. Botet Escriba a écrit :

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Le 01/06/13 12:34, Szymon Gatner a écrit :

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Hi Vincente,

I have found the error. My tests don't test for packaged_task.

The following patch should fix the issue (see attached file)

Please could you apply it a tell me if it fix with your compiler?

I am happy to confirm the fix, really appreciate so quick response from you. To clarify: I was using boost_1_54_0_beta1_rc1.7z from beta announcement.

Btw: I am using a workaround for packaged_task to be able store it in std/boost::function:

typedef boost::packaged_task Task; typedef boost::shared_ptr<Task> TaskPtr; TaskP task(new Task(f)); addTask(boost::bind(&Task::operator (), task));

where:

void addTask(boost::function task);

not a problem of course but with 1.54 beta I was still not able to just move packaged_task to function directly:

boost::packaged_task t1;

boost::function f2(boost::move(t1));

gives:

Error1error C2248: 'boost::packaged_task<<unnamed-symbol>>::packaged_task' : cannot access private member declared in class 'boost::packaged_task<<unnamed-symbol>>'d:\devel\boost_1_54_0_beta1\boost\function\function_template.hpp1070ConsoleApplication2

same error with std::function.

Is this an issue with packaged_task or a std/boost function implementations. I think boost::functions are not movable so they probably try to copy packaged_task but not sure why it also doesn't work with std::function.

IMO boost::function doesn't supports move semantics. I will take a look at std::function. Could you post the error with std::function?

I don't find a constructor of function from a rvalue functor.

From the standard: template<class F> function(F f); template function(allocator_arg_t, const A& a, F f); 7 Requires: F shall be *CopyConstructible*. f shall be Callable (20.10.11.2) for argument types ArgTypes and return type R. The copy constructor and destructor of A shall not throw exceptions. 8 Postconditions: !*this if any of the following hold: --- f is a NULL function pointer. --- f is a NULL pointer to member. --- F is an instance of the function class template, and !f 9 Otherwise, *this targets a copy of f initialized with std::move(f). [Note: Implementations are encouraged to avoid the use of dynamically allocated memory for small callable objects, for example, where f's target is an object holding only a pointer or reference to an object and a member function pointer. --- end note ] 10 Throws: shall not throw exceptions when f is a function pointer or a reference_wrapper<T> for some T. Otherwise, may throw bad_alloc or any exception thrown by F's copy or move constructor.

But I find an assignment

template<class F> function& operator=(F&& f); 18 Effects: function(std::forward<F>(f)).swap(*this); 19 Returns: *this

This seems confusing. What am I missing? I've a clear answer from Howard. See below.

Le 01/06/13 18:29, Howard Hinnant a écrit :

The reason F has to be CopyConstructible is not because of the signatures of the constructor and assignment operator. Both could be F, or F&&, and we appear to have split the difference. :-)

The reason F has to be CopyConstructible is because of function's copy constructor, and because F is type-erased inside of function. Because F is type-erased, when function is copied, it must copy F using a virtual function call (or via a function pointer). And because F's copy constructor must be called via a virtual function, F's copy constructor is instantiated whether or not function's copy constructor is instantiated. And that is why F must be CopyConstructible.

If F was not type-erased inside of function, F would not have to be CopyConstructible until actually copied.