(23.2.1 – 3)
For the components affected by this subclause that declare an allocator_type,objects stored in these components shall be constructed using the allocator_traits::construct function and destroyed using the allocator_traits::destroy function (20.6.8.2). These functions are called only for the container’s element type,not for internal types used by the container
(23.2.1 – 7)
Unless otherwise specified,all containers defined in this clause obtain memory using an allocator
是否真的,容器使用的所有内存都是由指定的分配器分配的?因为标准说内部类型不是用allocator_traits :: construct构造的,所以应该对operator new进行某种调用.但是标准也说这个子句中定义的所有容器都使用分配器来获取内存,在我看来这意味着它不能是普通的新运算符,它必须是放置新的运算符.我对么?
让我举个例子,为什么这很重要.
假设我们有一个类,它包含一些已分配的内存:
#include <unordered_map>
#include <iostream>
#include <cstdint>
#include <limits>
#include <memory>
#include <new>
class Arena
{
public:
Arena(std::size_t size)
{
size_ = size;
location_ = 0;
data_ = nullptr;
if(size_ > 0)
data_ = new(std::nothrow) uint8_t[size_];
}
Arena(const Arena& other) = delete;
~Arena()
{
if(data_ != nullptr)
delete[] data_;
}
Arena& operator =(const Arena& arena) = delete;
uint8_t* allocate(std::size_t size)
{
if(data_ == nullptr)
throw std::bad_alloc();
if((location_ + size) >= size_)
throw std::bad_alloc();
uint8_t* result = &data_[location_];
location_ += size;
return result;
}
void clear()
{
location_ = 0;
}
std::size_t getNumBytesUsed() const
{
return location_;
}
private:
uint8_t* data_;
std::size_t location_,size_;
};
我们也有自定义分配器:
template <class T> class FastAllocator
{
public:
typedef T value_type;
typedef T* pointer;
typedef const T* const_pointer;
typedef T& reference;
typedef const T& const_reference;
typedef std::size_t size_type;
typedef std::ptrdiff_t difference_type;
template <class U> class rebind
{
public:
typedef FastAllocator<U> other;
};
Arena* arena;
FastAllocator(Arena& arena_): arena(&arena_) {}
FastAllocator(const FastAllocator& other): arena(other.arena) {}
template <class U> FastAllocator(const FastAllocator<U>& other): arena(other.arena) {}
//------------------------------------------------------------------------------------
pointer allocate(size_type n,std::allocator<void>::const_pointer)
{
return allocate(n);
}
pointer allocate(size_type n)
{
return reinterpret_cast<pointer>(arena->allocate(n * sizeof(T)));
}
//------------------------------------------------------------------------------------
void deallocate(pointer,size_type) {}
//------------------------------------------------------------------------------------
size_type max_size() const
{
return std::numeric_limits<size_type>::max();
}
//------------------------------------------------------------------------------------
void construct(pointer p,const_reference val)
{
::new(static_cast<void*>(p)) T(val);
}
template <class U> void destroy(U* p)
{
p->~U();
}
};
这就是我们使用它的方式:
typedef std::unordered_map<uint32_t,uint32_t,std::hash<uint32_t>,std::equal_to<uint32_t>,FastAllocator<std::pair<uint32_t,uint32_t>>> FastUnorderedMap;
int main()
{
// Allocate memory in arena
Arena arena(1024 * 1024 * 50);
FastAllocator<uint32_t> allocator(arena);
FastAllocator<std::pair<uint32_t,uint32_t>> pairAllocator(arena);
FastAllocator<FastUnorderedMap> unorderedMapAllocator(arena);
FastUnorderedMaP* fastUnorderedMap = nullptr;
try
{
// allocate memory for unordered map
fastUnorderedMap = unorderedMapAllocator.allocate(1);
// construct unordered map
fastUnorderedMap =
new(reinterpret_cast<void*>(fastUnorderedMap)) FastUnorderedMap
(
0,std::hash<uint32_t>(),std::equal_to<uint32_t>(),pairAllocator
);
// insert something
for(uint32_t i = 0; i < 1000000; ++i)
fastUnorderedMap->insert(std::make_pair(i,i));
}
catch(std::bad_alloc badAlloc)
{
std::cout << "--- BAD ALLOC HAPPENED DURING FAST UnorDERED MAP INSERTION ---" << std::endl;
}
// no destructor of unordered map is called!!!!
return 0;
}
如您所见,unordered_map的析构函数永远不会被调用,但在破坏竞技场对象期间会释放内存.会有任何内存泄漏吗?为什么?
我真的很感激这个主题的任何帮助.
解决方法
> 2用于内存管理:分配/解除分配
> 2用于对象生存期管理:construct / destroy
引用中的这些函数仅适用于构造和销毁(在前一句中提到),而不是分配/解除分配,因此没有矛盾.
现在,关于内存泄漏,竞技场分配器不仅应该使用竞技场分配器(容器保证)来构建容器中的对象,而且还应该从该分配器获得这些对象分配的所有内存;不幸的是,这会变得稍微复杂一些.