我有以下问题:我使用我的项目自定义池Allocator,并且每次我要分配任何对象与位置新,析构函数也被称为隐含的对象。析构函数隐式调用,每次我分配放置新
这里是源代码:
测试对象:
class Obj {
public:
Obj(const std::string& s) {
std::cout << "Constructor Called" << std::endl;
}
~Obj() {
std::cout << "Destructor Called" << std::endl;
}
};
主营:
int main()
{
void *pmemory;
pmemory = malloc(ONEGIG_SIZE);
PoolAllocator* poolAllocator = new PoolAllocator(sizeof(Obj), __alignof(Obj), ONEGIG_SIZE, pmemory);
Obj *obj1 = allocator::allocateNew(*poolAllocator, Obj("Hello")); //<-- const and dest is called
//......
return 0;
}
这里是分配器的源代码: AllocateNew功能:
template <class T> T* allocateNew(Allocator& allocator, const T& t)
{
return new (allocator.allocate(sizeof(T), __alignof(T))) T(t);
}
池分配器:
PoolAllocator::PoolAllocator(size_t objectSize, u8 objectAlignment, size_t size, void* mem)
: Allocator(size, mem), _objectSize(objectSize), _objectAlignment(objectAlignment)
{
ASSERT(objectSize >= sizeof(void*));
//Calculate adjustment needed to keep object correctly aligned
u8 adjustment = pointer_math::alignForwardAdjustment(mem, objectAlignment);
_free_list = (void**)pointer_math::add(mem, adjustment);
size_t numObjects = (size-adjustment)/objectSize;
void** p = _free_list;
//Initialize free blocks list
for(size_t i = 0; i < numObjects-1; i++)
{
*p = pointer_math::add(p, objectSize);
p = (void**) *p;
}
*p = nullptr;
}
PoolAllocator::~PoolAllocator()
{
_free_list = nullptr;
}
void* PoolAllocator::allocate(size_t size, u8 alignment)
{
ASSERT(size == _objectSize && alignment == _objectAlignment);
if(_free_list == nullptr)
return nullptr;
void* p = _free_list;
_free_list = (void**)(*_free_list);
_used_memory += size;
_num_allocations++;
return p;
}
void PoolAllocator::deallocate(void* p)
{
*((void**)p) = _free_list;
_free_list = (void**)p;
_used_memory -= _objectSize;
_num_allocations--;
}
似乎在回复O P上之后在池分配器分配方法调用析构函数!任何人都可以解释为什么会发生?
我想说你应该添加调试打印* *所有*构造函数。有些东西告诉我,你看到的额外的析构函数实际上发生在一个临时任务上,可能在这里:'Obj * obj1 = allocator :: allocateNew(* poolAllocator,Obj(“Hello”));'。 – HolyBlackCat