如何使用C 11样式非默认构造分配器指定初始大小

在关于更新Visual Studio 2015 std::list::sort的先前线程中没有默认分配器。

我试图弄清楚如何创建具有初始(非零(大小的std::list实例，而无需在创建空列表后进行调整大小。

// this part of the code based on Microsoft example
template <class T>  
struct Mallocator  
{  
    typedef T value_type;  
    Mallocator(T) noexcept {} //default ctor not required by STL  
    // A converting copy constructor:  
    template<class U> Mallocator(const Mallocator<U>&) noexcept {}  
    template<class U> bool operator==(const Mallocator<U>&) const noexcept  
    {  
        return true;  
    }  
    template<class U> bool operator!=(const Mallocator<U>&) const noexcept  
    {  
        return false;  
    }  
    T* allocate(const size_t n) const;  
    void deallocate(T* const p, size_t) const noexcept;  
};  
template <class T>  
T* Mallocator<T>::allocate(const size_t n) const  
{
    if (n == 0)  
    {  
        return nullptr;  
    }  
    if (n > static_cast<size_t>(-1) / sizeof(T))  
    {  
        throw std::bad_array_new_length();  
    }  
    void* const pv = malloc(n * sizeof(T));  
    if (!pv) { throw std::bad_alloc(); }  
    return static_cast<T*>(pv);  
}  
template<class T>  
void Mallocator<T>::deallocate(T * const p, size_t) const noexcept  
{  
    free(p);  
}  
typedef unsigned long long uint64_t;
#define COUNT (4*1024*1024-1)   // number of values to sort
int main(int argc, char**argv)
{
    // this line from a prior answer
    // the (0) is needed to prevent compiler error, but changing the
    // (0) to (COUNT) or other non-zero value has no effect, the size == 0
    std::list <uint64_t, Mallocator<uint64_t>> ll(Mallocator<uint64_t>(0));
    // trying to avoid having to resize the list to get an initial size.
    ll.resize(COUNT, 0);