c++ - 使用模板容器的构造函数 C++ 时遇到问题

Question

我正在尝试使用自定义容器，并在该容器的构造函数中传递了一个内存池分配器。整个事情是这样开始的:

AllocatorFactory alloc_fac;

//Creates a CPool allocator instance with the size of the Object class
IAllocator* object_alloc = alloc_fac.GetAllocator<CPool>(1000,sizeof(Object));
//Creates a CPool allocator instance with the size of the BList<Object> class
IAllocator* list_alloc = alloc_fac.GetAllocator<CPool>(10,sizeof(BList<Object>));
//Same logic in here as well
IAllocator* node_alloc = alloc_fac.GetAllocator<CPool>(1000,sizeof(BListNode<Object>));

IAllocator 类如下所示:

 class IAllocator
{

public:

    virtual void* allocate( size_t bytes ) = 0;
    virtual void deallocate( void* ptr ) = 0;

    template <typename T>
    T* make_new()
    { return new ( allocate( sizeof(T) ) ) T (); }

    template <typename T, typename Arg0>
    T* make_new( Arg0& arg0 )
    { return new ( allocate( sizeof(T) ) ) T ( arg0 ); }

            .......
}

容器类的构造函数如下所示:

template <class T>
class BList {
......
public:
/**
 *@brief Constructor
 */
BList(Allocators::IAllocator& alloc){
     _alloc = alloc;
    reset();
    }
/**
 *@brief Constructor
 *@param inOther the original list
 */
 BList(const BList<T>& inOther){
    reset();
    append(inOther);
    }
.....
}

当我这样做时:

 BList<Object> *list = list_alloc->make_new<BList<Object>>(node_alloc);

编译器对此提示:

Error 1 error C2664: 'Containers::BList::BList(Allocators::IAllocator &)' : cannot convert parameter 1 from 'Allocators::IAllocator *' to 'Allocators::IAllocator &' c:\licenta\licenta-transfer_ro-02may-430722\licenta\framework\framework\iallocator.h 21 Framework

我想我对这个有点过头了....

Answer 1

现有的答案是正确的，但是关于如何自己阅读错误的快速说明:你只需要将它分成几部分......

Error 1 error C2664: 'Containers::BList::BList(Allocators::IAllocator &)' : cannot convert parameter 1 from 'Allocators::IAllocator *' to 'Allocators::IAllocator &'

阅读:

• 你正在调用 Containers::BList::BList(Allocators::IAllocator &)，它是一个构造函数，它接受一个参数，一个对 IAllocator 的引用>/li>
• cannot convert parameter 1 表示编译器在处理第一个(也是唯一一个)参数的类型时遇到问题
  • 你给了它这个类型:... from 'Allocators::IAllocator *'
  • 它需要这种类型(以匹配构造函数声明):... to 'Allocators::IAllocator &'

那么，如何将您拥有的指针转换为构造函数所需的引用？

---

好的，我也添加了实际答案:

Allocators::IAllocator *node_alloc = // ...
Allocators::IAllocator &node_alloc_ref = *node_alloc;
BList<Object> *list = list_alloc->make_new<BList<Object>>(node_alloc_ref);

或者只是:

BList<Object> *list = list_alloc->make_new<BList<Object>>(*node_alloc);