(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 分配器应该提供4个功能(这里感兴趣):> 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.
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