> For example, std::vector in C++ is a value type but the array storage is heap-allocated. This isn't correct. Firstly, C++ doesn't have value or reference typ

> For example, std::vector in C++ is a value type but the array storage is heap-allocated. This isn't correct. Firstly, C++ doesn't have value or reference types. Where any object is stored depends purely on how it is allocated. Either in the free store (dynamic) or automatic storage (the stack). std::vector has a constant object size - typically the size of 3 pointers (but that depends on the implementation - the standard doesn't specify): the start of the memory (dynamically allocated for storage), end/capacity (typically either another pointer or a size_t of the current capacity), and back (typically either a pointer just past the last element used or a size_t + 1 of current size). It's storage is dynamically allocated, though, and for a std::vector<T> will have dynamically allocated sizeof(T) * capacity bytes (maybe more due to alignment issues). std::array is constexpr safe and does not (directly) dynamically allocate memory - size has to be known at compile time (hence it's a template argument). For instance, on my compiler (GCC 4.8.4), for this example program: #include <iostream> #include <vector> #include <array> int main() { std::vector<int> v_ten(0, 10); std::vector<int> v_twenty(0, 20); std::array<int, 10> a_ten{0}; std::array<int, 20> a_twenty{0}; std::cout << "sizeof(v_ten): " << sizeof(v_ten) << std::endl; std::cout << "sizeof(v_twenty): " << sizeof(v_twenty) << std::endl; std::cout << "sizeof(a_ten): " << sizeof(a_ten) << std::endl; std::cout << "sizeof(a_twenty): " << sizeof(a_twenty) << std::endl; return 0; } produces this output: sizeof(v_ten): 24 sizeof(v_twenty): 24 sizeof(a_ten): 40 sizeof(a_twenty): 80 edit: added comment about alignment.