Does std::array<> guarantee allocation on the stack only?

Question

Is std::array<int,10> (without myself using new) guaranteed to be allocated in the stack rather then the heap by the C++-Standard?

To be clear, I do not mean new std::array<int, 10>. I mainly wonder, if the standard library is allowed to use new inside its implementation.

Answer 1

I could not find more explicit answer in the standard, but [array.overview]/2:

An array is an aggregate ([dcl.init.aggr]) that can be list-initialized with up to N elements whose types are convertible to T.

And [dcl.init.aggr]/1:

An aggregate is an array or a class (Clause [class]) with

• no user-provided, explicit, or inherited constructors ([class.ctor]),

...

That about covers it. No way an aggregate could allocate memory dynamically (or perhaps, do anything at all at its own during the construction). There's only an implicitly-declared trivial constructor.

^{Of course, if you new std::array<...>, you get an array on "the heap".}

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Some may be more satisfied by what we can get on cppreference:

std::array is a container that encapsulates fixed size arrays.

This container is an aggregate type with the same semantics as a struct holding a C-style array T[N] as its only non-static data member.

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Thirdly, std::array was introduced in C++11. Why? For example, to complement std::vector in some ways, like usage in constexpr functions, where dynamic allocation is not allowed.

Answer 2

TL;DR: yes, it is on the stack.

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The longer story:

C++ has no concept of stack or heap. Those are implementation details, and there is at least one platform that does not use a traditional stack (but rather linked list of heap allocations for it).

It has automatic storage and the free store. new accesses the free store, and variables "on the stack" go into automatic storage.

In practice, in order to allocate things on the free store, you have to risk an out of memory exception. So the general rule is things that guarantee they do not throw must be using automatic storage. array makes this guarantee (except whatever is in it can throw, naturally). It is also an aggregate of plain old data, effectively forced to look like:

template<class T,std::size_t N>
struct array {
  T __no_fixed_name__[N];
  // non-constructor/destructor methods omitted as they are noise at this point
};

In theory it could be implemented by the compiler via magic that is not actual C++, but there is no need for that, so nobody bothers.

So in conclusion: yes, std::array is on the stack.