使用自定义分配器但没有其他args的std :: function构造函数有什么意义？

Question

我正在使用std :: function和自定义分配器,但是当我没有为初始仿函数提供函数时,它的行为不像我预期的那样.

当我向构造函数提供自定义分配器但没有初始函子时,分配器从未使用过或者看起来如此.

这是我的代码.

//Simple functor class that is big to force allocations
struct Functor128
{
    Functor128()
    {}

    char someBytes[128];

    void operator()(int something)
    {
        cout << "Functor128 Called with value " << something << endl;
    }
};

int main(int argc, char* argv[])
{
Allocator<char, 1> myAllocator1;
Allocator<char, 2> myAllocator2;
Allocator<char, 3> myAllocator3;
Functor128 myFunctor;

cout << "setting up function1" << endl;
function<void(int)> myFunction1(allocator_arg, myAllocator1, myFunctor);
myFunction1(7);

cout << "setting up function2" << endl;
function<void(int)> myFunction2(allocator_arg, myAllocator2);
myFunction2 = myFunctor;
myFunction2(9);

cout << "setting up function3" << endl;
function<void(int)> myFunction3(allocator_arg, myAllocator3);
myFunction3 = myFunction1;
myFunction3(19);
}

输出:

setting up function1
Allocator 1 allocating 136 bytes.
Functor128 Called with value 7

setting up function2
Functor128 Called with value 9

setting up function3
Allocator 1 allocating 136 bytes.
Functor128 Called with value 19

所以case1:myFunction1按预期使用allocator1进行分配.

case2:myFunction2在构造函数中被赋予allocator2,但是当它被赋予一个仿函数时,它似乎重置为使用默认的std :: allocator来进行分配.(因此没有关于分配的打印).

case3:myFunction3在构造函数中被赋予allocator3但是当从myFunction1赋值时,使用function1的allocator进行分配以进行分配.

这是正确的行为吗？特别是,在案例2中为什么要恢复使用默认的std :: allocator？如果是这样,那么采用分配器的空构造函数的重点是什么,因为分配器永远不会被使用.

我正在使用VS2013代码.

我的Allocator类只是一个最小的实现,它使用new并在分配时注销

template<typename T, int id = 1>
class Allocator {
public:
    //    typedefs
    typedef T value_type;
    typedef value_type* pointer;
    typedef const value_type* const_pointer;
    typedef value_type& reference;
    typedef const value_type& const_reference;
    typedef std::size_t size_type;
    typedef std::ptrdiff_t difference_type;

public:
    //    convert an allocator<T> to allocator<U>
    template<typename U>
    struct rebind {
        typedef Allocator<U> other;
    };

public:
    inline  Allocator() {}
    inline ~Allocator() {}
    inline  Allocator(Allocator const&) {}
    template<typename U>
    inline  Allocator(Allocator<U> const&) {}

    //    address
    inline pointer address(reference r) { return &r; }
    inline const_pointer address(const_reference r) { return &r; }

    //    memory allocation
    inline pointer allocate(size_type cnt,
        typename std::allocator<void>::const_pointer = 0) 
    {
        size_t numBytes = cnt * sizeof (T);
        std::cout << "Allocator " << id <<  " allocating " << numBytes << " bytes." << std::endl;
        return reinterpret_cast<pointer>(::operator new(numBytes));
    }
    inline void deallocate(pointer p, size_type) {
        ::operator delete(p);
    }

    //    size
    inline size_type max_size() const {
        return std::numeric_limits<size_type>::max() / sizeof(T);
    }

    //    construction/destruction
    inline void construct(pointer p, const T& t) { new(p)T(t); }
    inline void destroy(pointer p) { p->~T(); }

    inline bool operator==(Allocator const&) { return true; }
    inline bool operator!=(Allocator const& a) { return !operator==(a); }
};    //    end of class Allocator

Answer 1

std::function分配器支持是......很奇怪.

目前的规格operator=(F&& f)是它确实如此std::function(std::forward<F>(f)).swap(*this);.如您所见,这意味着默认情况下f使用任何std::function使用来分配内存,而不是用于构造的分配器*this.所以你观察到的行为是正确的,尽管令人惊讶.

此外,由于(allocator_arg_t, Allocator)和(allocator_arg_t, Allocator, nullptr_t)构造函数是noexcept,它们甚至不能真正存储分配器,即使它们想要(类型擦除分配器可能需要动态分配).因此,它们基本上是无操作,以支持使用分配器构造协议.

LWG最近拒绝了一个会改变这种行为的问题.