The constexpr array size problem

on under c++
13 minute read

This issue was first pointed out to me by Michael Park, and mostly explained to me by T.C.

Let’s say I have an array, and I want to get its size and use it as a constant expression. In C, we would write a macro for this:

#define ARRAY_SIZE(a) (sizeof(a)/sizeof(a[0]))

Macro notwithstanding, this works fine in all contexts. In C++, it unfortunately also compiles for any types with overloaded operator[] and gives a nonsense result. Can we provide a type-safe way to do better?

We have constexpr, so let’s use it:

template <typename T, size_t N>
constexpr size_t array_size(T (&)[N]) {
    return N;

This beats the C macro approach both by not being a macro and by not giving bogus answers for vector<T>. But it has possibly-surprising limitations:

void check(int const (&param)[3]) {
    int local[] = {1, 2, 3};
    constexpr auto s0 = array_size(local); // ok
    constexpr auto s1 = array_size(param); // error

All compilers reject s1 (gcc still accepted it when I started writing this post, but that bug has been fixed. That’s some timing!)

Wait, why?

All of these compilers are actually correct to reject this example. The reason is that in order for array_size(param) to work, we have to pass that reference to param into array_size - and that involves “reading” the reference. The specific rule we’re violating is [expr.const]/4.12. The reason we violate the reference rule is due to the underlying principle that the constant evaluator has to reject all undefined behavior (UB is a compile error during constant evaluation!) and so the compiler has to check that all references are valid.

This would be more obvious if our situation used pointers instead of references:

template <typename T, size_t N>
constexpr size_t array_size(T (*)[N]) {
    return N;

void check(int const (*param)[3]) {
    constexpr auto s2 = array_size(param); // error

This case has to be ill-formed, copying a function parameter during constant evaluation means it has to itself be a constant expression, and function parameters are not constant expressions - even in constexpr or consteval functions.

But if the param case is ill-formed, why does the local case work? An unsatisfying answer is that… there just isn’t any rule in [expr.const] that we’re violating. There’s no lvalue-to-rvalue conversion (we’re not reading through the reference in any way yet) and we’re not referring to a reference (that’s the previous rule we ran afoul of). With the param case, the compiler cannot know whether the reference is valid, so it must reject. With the local case, the compiler can see for sure that the reference to local would be a valid reference, so it’s happy.

Notably, the rule we’re violating is only about references. We can’t write a function that takes an array by value, so let’s use the next-best thing: std::array and use the standard library’s std::size (cppref):

void check_arr_val(std::array<int, 3> const param) {
    std::array<int, 3> local = {1, 2, 3};
    constexpr auto s3 = std::size(local); // ok
    constexpr auto s4 = std::size(param); // ok

If param were a reference, the initialization of s4 would be ill-formed (for the same reason as previously), but because it’s a value, this is totally fine.

So as long as you pass all your containers around by value, you’re able to use get and use the size as a constant expression. Which is the kind of thing that’s intellectually interesting, but also wildly impractical because obviously nobody’s about to start passing all their containers around by value.

Why might we care?

Before getting into more detail about the problem itself, let’s take a look at why this matters. The following is just one motivating example. I picked it because it’s probably the easiest to look at, but just keep in mind that this is far from the only reason we might care about this sort of thing.

C++20 will have a new type std::span (cppref, I’ve written about span before here), which is a contiguous view on T. span comes in two flavors: dynamic extent and fixed extent. Roughly speaking:

// fixed-extent: always has size Extent, only needs
// to store a single pointer
template <typename T, size_t Extent = dynamic_extent>
struct span {
    T* ptr;
    constexpr T* begin() { return ptr; }
    constexpr T* end() { return ptr + Extent; }
    constexpr size_t size() const { return Extent; }

// dynamic-extent: size is variable, needs to store
// both pointer and size
template <typename T>
struct span<T, dynamic_extent> {
    T* ptr;
    size_t size;
    constexpr T* begin() const { return ptr; }
    constexpr T* end() const { return ptr + size; }
    constexpr size_t size() const { return size; }

It’s not a complex type.

One of the big features of span<T> is that all contiguous ranges over T are convertible to it. This conversion is safe, cheap, and desirable.

But that isn’t necessarily the case for span<T, 5> - we certainly don’t want that to be implicitly constructible from vector<T>, since how do we know if the incoming vector has enough elements in it? The direction we’re going (see P1976) is that this constructor will instead be explicit. That is:

void f(std::span<int, 5>);

std::vector<int> v3 = {1, 2, 3};
f(v3);                    // ill-formed
f(std::span<int, 5>(v3)); // compiles but UB

std::vector<int> v5 = {1, 2, 3, 4, 5};
f(std::span<int, 5>(v5)); // well-formed

But even with fixed-extent, there are some conversions that would be perfectly safe to be implicit: arrays! Arrays have the size encoded in the type, we surely know at compile time if an array has 5 elements or not, so this could be perfectly fine:

void f(std::span<int, 5>);

int elems[] = {1, 2, 3, 4, 5};
f(elems); // perfectly safe, known statically

How could we make this work?

We’d have to make the constraint something like the following (other constraints omitted for brevity, like making sure the range is actually contiguous and having the right underlying type):

template <typename T, size_t Extent>
struct span {
    template <range R>
    span(R&& r)
        requires (std::size(r) == Extent);

As already pointing out, since r is a reference, std::size(r) cannot be a constant expression, so this constraint cannot be made to work. This would work fine if r was not a reference (but then we’d be constructing a span to refer to a range that is about to get destroyed, so this is a particularly tortured use of “work fine”).

Is there a library solution for this?

In the case I’m describing here, the size is encoded in the type. So if we change the way we query the size to query based on the type of the object instead of the value of the object, we can side-step these problems:

template <typename T> struct type_t { using type = T; };
template <typename T> inline constexpr type_t<T> type{};

template <typename T, size_t N>
constexpr size_t type_size(type_t<T[N]>) {
    return N;

template <typename T>
constexpr auto type_size(type_t<T>) -> decltype(T::size()) {
    return T::size();

template <typename T, size_t Extent>
struct span {
    template <range R>
        requires (type_size(type<std::remove_cvref_t<R>>)
                  == Extent);
    span(R&& r)

This works for both C arrays (the first overload) and std::array (the second). A similar approach was proposed in P1419, the only difference was that the proposal spelled this approach static_extent_v<std::remove_cvref_t<R>> instead of type_size(type<std::remove_cvref_t<R>>).

But this can only work for a range whose size is encoded into its type. It can’t work for a range whose size is a constant expression - which is at least conceptually what we want. One such range? Why span, of course:

constexpr int arr[] = {1, 2, 3, 4, 5};
constexpr std::span<int> s = arr; 

// this is fine, s.size() is a constant expression
static_assert(s.size() == 5);

// ... but this still wouldn't work!
std::span<int, 5> fixed = s;

s’s size is a constant expression, but it’s not tied to its type - these type- based approaches wouldn’t work.

So it’s, at best, a partial and unsatisfying solution. But at least it does offer a way to reliably get the size of an array as a constant expression - it’s just that we have to go back to using a macro:

#define ARRAY_SIZE(a) type_size( \

All this constexpr machinery, and the best we can do is really only a little bit better than C.

Is there a language solution for this?

There is no library solution for this, maybe there’s a language one? The obvious language rule would be to say something like: simply passing around a reference from one function to another is fine - it’s not until you actually need the reference for something that we start requiring it to be a valid constant expression. That is, we have to read through the reference in some way or do some operation that requires the address to be known, something in that vein.

A different, less generous, way to describe a feature like this would be to say that x.f() can still be a constant expression even if x is a reference to unknown object, as long as f is a static member function and x is a sufficiently direct naming of the reference (for some as-yet defined definition of sufficiently direct). This brings up other cases that would need to be considered (courtesy of Richard Smith):

  • (*ptr_to_array)->size()?
  • (*ptrs_to_arrays[3])->size()? What about if ptrs_to_arrays was actually an array of only 2 arrays, at which point we’d surely want to reject this during constant evaluation in the same way we reject out-of-bounds access generally?

And if we do still do bounds checking in the latter case, is there actually a line we can draw for which things we do on the left-hand-side (like this bounds-check) but which things we don’t (like actually require that *ptrs_to_arrays[3] is a constant expression itself)?

Moreover, even if we go this direction. Let’s say that we come up with some kind of exception in this space, such that the original code we tried to write actually works. Just reproducing the relevant code for proximity:

template <typename T, size_t Extent>
struct span {
    template <range R>
    span(R&& r)
        requires (std::size(r) == Extent);

void f(span<int, 5>);

int c_array[5];
std::array<int, 5> cpp_array;

f(c_array);   // now ok
f(cpp_array); // now ok

Would this even be a sufficient rule? Still no! It still wouldn’t work for the example I showed earlier:

constexpr std::array<int, 5> some_const_array = {1, 2, 3, 4, 5};
constexpr std::span<int> dynamic_span = some_const_array; // ok
static_assert(dynamic_span.size() == 5); // ok

f(dynamic_span); // still error!

Even if we say that we can propagate references to our heart’s delight during constant evaluation, as long as we don’t read them, that wouldn’t help us here. dynamic_span is a dynamic span, its size() member function is non-static, so we very much do need to read through the reference.

In order to make this example work, we’d need to not only have a reference propagation rule (to make the c_array and cpp_array cases work), but we’d also need to adopt something like function parameter constraints (see P1733 and P2049, and my response D2089) or, better, constexpr function parameters (see P1045). Either way, these proposals would only help the dynamic_span case - with the array cases, since the arrays themselves aren’t constant expressions, none of what they are suggesting would help. We’d need both.

The emplace_back() problem

But even then, we’re still stuck. We still have what I’m calling the vector push_back() / emplace_back() problem (I first pointed this out in D2089). How can we make all of these work:

std::vector<std::span<int, 5>> bunch_of_spans;


The three calls to push_back all invoke a function whose signature would be:

void push_back(std::span<int, 5> const&);

That is, the conversion to span happens on the way into the function. For dynamic_span, it’s still a constant expression here - so either we can still treat it as a constant expression for constraint purposes (as with the function parameter constraints approach in P1733) or we can add an overloaded constructor to span that takes a constexpr range (as with constexpr parameter approach in P1045). That part works fine.

But the three calls to emplace_back() all invoke something that is roughly equivalent to (this isn’t exactly correct, but for the purposes of this discussion, it’s good enough):

template <typename Arg>
void emplace_back(Arg&& arg) {
    push_back(std::span<int, 5>(std::forward<Arg>(Arg));

That is, the conversion so span happens inside of emplace_back(). In order for this to still work for dynamic_span, we would need to somehow remember that arg refers to a constant expression. And while dynamic_span is a constant expression, bunch_of_spans doesn’t have to be - this could be runtime code. How can this runtime call remember the “constant-expression-ness” of its parameter? Ideally without having to touch std::vector<T>::emplace_back in any way whatsoever?

I have no idea.

But I would love to get to the point where this code actually compiles:

constexpr int c_array[] = {1, 2, 3, 4, 5};
constexpr std::span<int> dynamic_span = c_array;

std::vector<std::span<int, 5>> bunch_of_spans;


Basically, the best way to get the size of an array to be used as a constant expression is still to use a macro - in C++, we can make that macro more type safe than the initial C version, but still a macro.

Getting to the point where we can access the size of a range as a constant expression - whether that size is part of the type (as it is for C arrays and std::array) or a variable part of a constexpr object (as it would be for a wide variety of ranges) - would require multiple language changes.

And none of the hypothetical language changes I’ve described in this post are exactly trivial either, so I expect we’ll have to live this problem for a while…