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Template Delete

SajberToffe · 2003-10-23T07:50:04

Hello! I have some plans for a engine. After reading through superpig''s great articles I figured that I''d need some kind of MemoryManagemanet. I didn''t want every object to have a common baseclass, rather that any created class without modification would be useable. I thought refcounting would be easiest to implement. After trial and error, I came up with this. template<typename T> struct Node { T data; Node<T> * next; }; inline void * operator new(size_t size) { if(!size)size++; Node<unsigned int> * node = (Node<unsigned int> *)malloc(size + sizeof(Node<unsigned int>)); node->data = 0; // References node->next = Registry::headNode; Registry::headNode = node; return (void*)((int)node + sizeof(Node<unsigned int>)); } template<typename T> void Delete(T * data) { Node<unsigned int> * ptr = (Node<unsigned int> *)((int)data - sizeof(Node<unsigned int>)); data->~T(); free(ptr); //free(data) screws the memory up } void AddRef(void * ptr) { Node<unsigned int> * node = (Node<unsigned int> *)((int)ptr - sizeof(Node<unsigned int>)); node->data++; } void Release(void * ptr) { Node<unsigned int> * node = (Node<unsigned int> *)((int)ptr - sizeof(Node<unsigned int>)); node->data--; } The objects would all be freed from some Registry class, no delete calls would be allowed. Now my questions are: (1)Are objects really freed when using free() on the ptr pointer? I know that the destructor is called, but is the whole memory block made free? (2)Is 2 byte too expensive for every object? There are some trouble when pointing to a arrayed object, but still... Sorry for a confusing post

General and Gameplay Programming Programming

Started by SajberToffe October 22, 2003 04:52 PM

11 comments, last by SajberToffe 20 years, 5 months ago

Anonymous

October 23, 2003 04:17 AM

actually pointer sizes can vary altough they can be at most sizeof(void *) * CHAR_BITS / 8 bytes.

amag

152

October 23, 2003 06:53 AM

quote:
...cause one can't get the size from delete(void * ptr)...

Actually you can, just declare you delete-operator as

void operator delete(void *p, std::size_t size) {}

more people than you have found the need for size to be an argument of delete...

[edited by - amag on October 23, 2003 7:54:50 AM]

SajberToffe

127

Author

October 23, 2003 07:50 AM

That''s true... maby it''s not much of a difference anyway... perhaps I''ll stick with operator delete...

But still, I don''t know someone answered my main question. If the whole memory block is freed when freeing from the node part.

Template Delete

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