monotonic::storage<100> store; { typedef std::vector<char, monotonic::allocator<char> > Vector; Vector vec(store); vec.resize(10); // on the stack vec.resize(1000); // now on the heap }
I am telling you, this code is not portable. Some compilers will create code that silently crashes because std::vector is not required to use any specific monotonic::allocator. Please, please remember that it is REQUIRED that all allocators of the same type are completely interchangeable in STL containers. That means NO non-static data. So what you have written below could fail on some compilers. This type of failure is especially noticable in lists, and almost unavoidable when using the splice member function if the compiler makes the allocator equality assumption. Actually, in general, I would assume this code would fail: monotonic::storage<100> store1; boost::monotonic::list<int> list1(store1); { monotonic::storage<100> store2; boost::monotonic::list<int> list2(store2); list2.push_back(1); list1.splice(list1.begin(), list2); } std::cout << *list1.begin(); // segfault? Though I haven't tested it. Of course, I don't think there's a way to fix that. But it's something to know. Also, your allocator's member pointer storage is public. I assume it's supposed to be private?