Why is std::array< T, 0 > not empty?

Question

Given any std::array< T, 0 >, why is it not empty? I mean "empty" as in:

std::is_empty< std::array< int, 0 > >::value

... returning false, and

#include <iostream>
#include <tuple>
#include <array>

struct Empty {};

int main()
{
    std::cout << sizeof(std::tuple<int>) << std::endl;
    std::cout << sizeof(std::tuple<int,Empty>) << std::endl;
    std::cout << sizeof(std::tuple<int,std::array<int,0>>) << std::endl;
}

yields

4
4
8

This means that for std::array<int,0>, the empty base optimization (EBO) is not applied.

This seem especially strange to me given that std::tuple<> (note: no template parameters) is empty, i.e., std::is_empty<std::tuple<>>::value does yield true.

Question: Why is that, given that size 0 is already a special case for std::array? Is it intentional or an oversight in the standard?

Jonathan Wakely · Accepted Answer · 2013-03-20 00:29:59Z

28

The standard doesn't say anything about whether tuple or array should be empty, what you're seeing are implementation details, but there's no reason to make tuple<> non-empty, whereas there is a good reason for array<T, 0> being non-empty, consider:

std::array<int, sizeof...(values)> = { { values... } };

When the parameter pack is empty you'd get:

std::array<int, 0> = { { } };

For the initializer to be valid the object needs a member, which cannot be int[0] because you can't have zero-sized arrays as members, so a possible implementation is int[1]

An implementation doesn't have to special case the whole array, it can just do:

T m_data[N == 0 ? 1 : N];

and all other members work exactly the same way (assuming end() is defined as begin()+N)

edited Mar 20, 2013 at 0:29

answered Mar 20, 2013 at 0:10

Jonathan Wakely

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15 Comments

Daniel Frey Over a year ago

GCC 4.8 seems to do it differently (or the libstdc++ coming with it), since sizeof(std::array<int,0>) == 1. But I realize you said "one possible implementation" and I accept the reasoning, so thanks!

Jonathan Wakely Over a year ago

GCC does value_type _M_instance[_Nm ? _Nm : 1]; and I see sizeof(array<int,0>) == sizeof(int)

Daniel Frey Over a year ago

I see sizeof(std::array<int,0>) == 1 for GCC 4.8 on LiveWorkSpace.org. Strange.

Jonathan Wakely Over a year ago

Ah yes, I forgot it changed for GCC 4.8 to allow zero-sized arrays of non-DefaultConstructible types, see gcc.gnu.org/PR53248 (when gcc.gnu.org is back online). It now has an empty struct member instead of value_type[1].

Daniel Frey Over a year ago

I just found §23.3.2.1/2 from the standard. Doesn't that mean that it's not just an implementation detail and that the standard does say something about std::array<T,0>'s emptiness (although indirectly)? As a consequence, if I read it correctly, an empty implementation would even be illegal, wouldn't it?

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