Question

我从DirectX获取了存储顶点信息的存储位置。处理顶点信息的一种非常方便的方法是使用std :: vector＆lt;＆gt;包含顶点信息的结构。

鉴于我有一个指向大缓冲区的指针，我可以使用std :: vector来管理缓冲区中的元素吗？经常构造一个std :: vector使它有自己的地址，这不是我想要的。我可以以某种方式使用操作员放置吗？

Answer 1

是的，你可以。使用custom allocator。在此DirectX内存的分配器返回地址中。

以下是基于Compelling examples of custom C++ STL allocators?答案的完整考试。此解决方案在分配器中使用placement new。

#include <memory>
#include <iostream>
#include <vector>

using namespace std;

template <typename T>
class placement_memory_allocator: public std::allocator<T>
{
    void* pre_allocated_memory;
public:
    typedef size_t size_type;
    typedef T* pointer;
    typedef const T* const_pointer;

    template<typename _Tp1>
    struct rebind
    {
            typedef placement_memory_allocator<_Tp1> other;
    };

    pointer allocate(size_type n, const void *hint=0)
    {
            char* p = new(pre_allocated_memory)char[n * sizeof(T)];
            cout << "Alloc " << n * sizeof(T) << " bytes @" << hex << (void*)p <<endl;
            return (T*)p;
    }

    void deallocate(pointer p, size_type n)
    {
            cout << "Dealloc " << n << " bytes @" << hex << p << endl;
            //delete p;
    }

    placement_memory_allocator(void* p = 0) throw(): std::allocator<T>(), pre_allocated_memory(p) { cout << "Hello allocator!" << endl; }
    placement_memory_allocator(const placement_memory_allocator &a) throw(): std::allocator<T>(a) {pre_allocated_memory = a.pre_allocated_memory;}
    ~placement_memory_allocator() throw() { }
};

class MyClass
{   
    char empty[10];
    char* name;
public:
    MyClass(char* n) : name(n){ cout << "MyClass: " << name << " @" << hex << (void*)this << endl; }
    MyClass(const MyClass& s){ name = s.name; cout << "=MyClass: " << s.name << " @" << hex << (void*)this << endl; }
    ~MyClass(){ cout << "~MyClass: " << name << " @" << hex << (void*)this <<  endl; }
};

int main()
{
    // create allocator object, intialized with DirectX memory ptr.
    placement_memory_allocator<MyClass> pl(DIRECT_X_MEMORY_PTR);
    //Create vector object, which use the created allocator object.
    vector<MyClass, placement_memory_allocator<MyClass>> v(pl);
    // !important! reserve all the memory of directx buffer.
    // try to comment this line and rerun to see the difference
    v.reserve( DIRECT_X_MEMORY_SIZE_IN_BYTES / sizeof(MyClass));

    //some push backs.
    v.push_back(MyClass("first"));
    cout << "Done1" << endl;
    v.push_back(MyClass("second"));
    cout << "Done1" << endl;
    v.push_back(MyClass("third"));
    cout << "Done1" << endl;

}

Answer 2

std::vector的元素在堆上动态分配（new本身为std::vector），使得它们在内存中始终是连续的。因此，如果您的结构是vertex，那么使用std::vector<vertex>并不是您想要的，因为向量的元素不会位于您的大缓冲区中。

但是，您可以使用std::vector<vertex*>，例如：

vertex* bigBuffer;     // provided by DirectX
size_t bigBufferLen;   // provided by DirectX

std::vector<vertex*> array;
for (size_t i = 0; i < bigBufferLen; ++i)
{
    array.push_back(bigBuffer + i * sizeof(vertex));
}

我可以使用std :: vector作为预分配（原始）数组的外观吗？

2 个答案: