13

这个问题是关于在插入 std::map 期间自定义分配器实例的构造。

这是一个自定义分配器std::map<int,int>以及一个使用它的小程序:

#include <stddef.h>
#include <stdio.h>
#include <map>
#include <typeinfo>

class MyPool {
public:
  void * GetNext() {
    return malloc(24);
  }
  void Free(void *ptr) {
    free(ptr);
  }
};

template<typename T>
class MyPoolAlloc {
public:
  static MyPool *pMyPool;

  typedef size_t     size_type;
  typedef ptrdiff_t  difference_type;
  typedef T*         pointer;
  typedef const T*   const_pointer;
  typedef T&         reference;
  typedef const T&   const_reference;
  typedef T          value_type;

  template<typename X>
  struct rebind
  { typedef MyPoolAlloc<X> other; };

  MyPoolAlloc() throw() {
    printf("-------Alloc--CONSTRUCTOR--------%08x %32s\n", this, typeid(T).name());
  }

  MyPoolAlloc(const MyPoolAlloc&) throw()  {
    printf(" Copy Constructor ---------------%08x %32s\n", this, typeid(T).name());
  }

  template<typename X>
  MyPoolAlloc(const MyPoolAlloc<X>&) throw() {
    printf(" Construct T Alloc from X Alloc--%08x %32s %32s\n", this, typeid(T).name(), typeid(X).name());
  }

  ~MyPoolAlloc() throw() {
    printf(" Destructor ---------------------%08x %32s\n", this, typeid(T).name());
  };

  pointer address(reference __x) const { return &__x; }

  const_pointer address(const_reference __x) const { return &__x; }

  pointer allocate(size_type __n, const void * hint = 0) {
    if (__n != 1)
      perror("MyPoolAlloc::allocate: __n is not 1.\n");
    if (NULL == pMyPool) {
      pMyPool = new MyPool();
      printf("======>Creating a new pool object.\n");
    }
    return reinterpret_cast<T*>(pMyPool->GetNext());
  }

  //__p is not permitted to be a null pointer
  void deallocate(pointer __p, size_type __n) {
    pMyPool->Free(reinterpret_cast<void *>(__p));
  }

  size_type max_size() const throw() {
    return size_t(-1) / sizeof(T);
  }

  void construct(pointer __p, const T& __val) {
    printf("+++++++ %08x %s.\n", __p, typeid(T).name());
    ::new(__p) T(__val);
  }

  void destroy(pointer __p) {
    printf("-+-+-+- %08x.\n", __p);
    __p->~T();
  }
};

template<typename T>
inline bool operator==(const MyPoolAlloc<T>&, const MyPoolAlloc<T>&) {
  return true;
}

template<typename T>
inline bool operator!=(const MyPoolAlloc<T>&, const MyPoolAlloc<T>&) {
  return false;
}

template<typename T>
MyPool* MyPoolAlloc<T>::pMyPool = NULL;

int main(int argc, char *argv[]) {

  std::map<int, int, std::less<int>, MyPoolAlloc<std::pair<const int,int> > > m;
  //random insertions in the map
  m.insert(std::pair<int,int>(1,2));
  m[5] = 7;
  m[8] = 11;
  printf("======>End of map insertions.\n");
  return 0;
}

这是该程序的输出:

--------Alloc--CONSTRUCTOR--------bffcdaa6 St4pairIKiiE
 从 X Alloc 构造 T Alloc--bffcda77 St13_Rb_tree_nodeISt4pairIKiiEE St4pairIKiiE
 复制构造函数 ----------------bffcdad8 St13_Rb_tree_nodeISt4pairIKiiEE
 析构函数 ---------bffcda77 St13_Rb_tree_nodeISt4pairIKiiEE
 析构函数---------bffcdaa6 St4pairIKiiE
======>创建一个新的池对象。
 从 X Alloc 构造 T Alloc--bffcd9df St4pairIKiiE St13_Rb_tree_nodeISt4pairIKiiEE
+++++++ 0985d028 St4pairIKiiE。
 析构函数 ---------------------bffcd9df St4pairIKiiE
 从 X Alloc 构造 T Alloc--bffcd95f St4pairIKiiE St13_Rb_tree_nodeISt4pairIKiiEE
+++++++ 0985d048 St4pairIKiiE。
 析构函数 ---------------------bffcd95f St4pairIKiiE
 从 X Alloc 构造 T Alloc--bffcd95f St4pairIKiiE St13_Rb_tree_nodeISt4pairIKiiEE
+++++++ 0985d068 St4pairIKiiE。
 析构函数 ---------------------bffcd95f St4pairIKiiE
======>地图插入结束。
 从 X Alloc 构造 T Alloc--bffcda23 St4pairIKiiE St13_Rb_tree_nodeISt4pairIKiiEE
-+-+-+- 0985d068。
 析构函数---------bffcda23 St4pairIKiiE
 从 X Alloc 构造 T Alloc--bffcda43 St4pairIKiiE St13_Rb_tree_nodeISt4pairIKiiEE
-+-+-+- 0985d048。
 析构函数---------bffcda43 St4pairIKiiE
 从 X Alloc 构造 T Alloc--bffcda43 St4pairIKiiE St13_Rb_tree_nodeISt4pairIKiiEE
-+-+-+- 0985d028。
 析构函数---------bffcda43 St4pairIKiiE
 析构函数 ---------bffcdad8 St13_Rb_tree_nodeISt4pairIKiiEE

输出的最后两列显示,std::pair<const int, int>每次插入映射时都会构造一个分配器。为什么这是必要的?有没有办法抑制这种情况?

谢谢!

编辑:此代码在 g++ 版本 4.1.2 的 x86 机器上测试。如果您希望在 64 位机器上运行它,您至少必须更改return malloc(24). 更改为return malloc(48)应该工作。

4

2 回答 2

1

之所以如此,是因为分配器是 forstd::pair<const int, int>但实现实际上需要分配一个更复杂的数据结构,它是其中的一个成员。虽然我希望需要构造和缓存实际的分配器,但每次都重新构造它并不违法。这是一个实现细节,如果不更改实现,您就无法逃避。创建的实际分配器类型是St13_Rb_tree_nodeISt4pairIKiiEE(mangled name)。

于 2012-07-07T09:42:05.650 回答
1

在 MyPool.h(单例)中:

class MyPool
{
...
public:
  static MyPool & GetInstance( void );
private:
  MyPool(void);
}

在 MyPool.cpp 中:

MyPool & MyPool::GetInstance( void )
{
  static MyPool retval;
  return retval;
}

在 fooStdAllocator.h 中:

#pragma once

#include "MyPool.h"

#pragma push_macro( "new" )
#undef new
#include <new>

template <class T1> class fooStdAllocator;

// Description:
// Specialize for void
template <> class fooStdAllocator<void>
{
public:
  typedef void * pointer;
  typedef const void* const_pointer;
  typedef void value_type;
  template <class U1> struct rebind { typedef fooStdAllocator<U1> other; };
};

template <class T1> class fooStdAllocator
{
public:
  // Description:
  // Typedefs
  typedef T1 value_type;
  typedef size_t size_type;
  typedef ptrdiff_t difference_type;
  typedef T1* pointer;
  typedef const T1* const_pointer;
  typedef T1& reference;
  typedef const T1& const_reference;

  // Description:
  // The rebind member allows a container to construct an allocator for some arbitrary type out of
  // the allocator type provided as a template parameter.
  template <class U1> struct rebind { typedef fooStdAllocator<U1> other; };

  // Description:
  // Constructors
  fooStdAllocator( void ) : pool(MyPool::GetInstance()) {};
  fooStdAllocator( const fooStdAllocator& other ) : pool(MyPool::GetInstance()) {};
  template <class U1> fooStdAllocator(const fooStdAllocator<U1>&) : pool(MyPool::GetInstance()) {};

  // Description:
  // Destructor
  ~fooStdAllocator( void ) {};

  // Description:
  // Returns the address of r as a pointer type. This function and the following function are used
  // to convert references to pointers.
  pointer address(reference r) const { return &r; };
  const_pointer address(const_reference r) const { return &r; };

  // Description:
  // Allocate storage for n values of T1.
  pointer allocate( size_type n, fooStdAllocator<void>::const_pointer hint = 0 )
  {
    // I would never do it that way:
    //pointer return_value = reinterpret_cast<pointer>( pool.GetNext() );
    // I would prefer to use the got size to allocate:
    pointer return_value = reinterpret_cast<pointer>( pool.GetNext(n) );

    if ( return_value == 0 )
      throw std::bad_alloc();
    return return_value;
  };

  // Description:
  // Deallocate storage obtained by a call to allocate.
  void deallocate(pointer p, size_type n)
  {
    pool.Free(p);
  };

  // Description:
  // Return the largest possible storage available through a call to allocate.
  size_type max_size() const
  {
    size_type return_value = 0xFFFFFFFF;
    return_value /= sizeof(T1);
    return return_value;
  };

  // Description:
  // Construct an object of type T1 at the location of ptr
  void construct(pointer ptr)
  {
    ::new (reinterpret_cast<void*>(ptr)) T1;
  };

  // Description:
  // Construct an object of type T1 at the location of ptr, using the value of U1 in the call to the
  // constructor for T1.
  template <class U1> void construct(pointer ptr, const U1& val)
  {
    ::new (reinterpret_cast<void*>(ptr)) T1(val);
  };

  // Description:
  // Construct an object of type T1 at the location of ptr, using the value of T1 in the call to the
  // constructor for T1.
  void construct(pointer ptr, const T1& val)
  {
    ::new (reinterpret_cast<void*>(ptr)) T1(val);
  };

  // Description:
  // Call the destructor on the value pointed to by p
  void destroy(pointer p)
  {
    p->T1::~T1();
  };
private:
  MyPool &pool;
};

// Return true if allocators b and a can be safely interchanged. "Safely interchanged" means that b could be
// used to deallocate storage obtained through a and vice versa.
template <class T1, class T2> bool operator == ( const fooStdAllocator<T1>& a, const fooStdAllocator<T2>& b)
{
  return true;
};
// Return false if allocators b and a can be safely interchanged. "Safely interchanged" means that b could be
// used to deallocate storage obtained through a and vice versa.
template <class T1, class T2> bool operator != ( const fooStdAllocator<T1>& a, const fooStdAllocator<T2>& b)
{
  return false;
};
#pragma pop_macro( "new" )

您可以按如下方式使用它:

std::map<keyT,valueT,std::less<keyT>,fooStdAllocator> your_map;
于 2012-07-07T15:10:39.753 回答