Must T be unqualified in std::allocator<T>?

Question

gcc, clang, and msvc all reject the following code:

#include <memory>
#include <vector>

int main() {
    auto _ = std::vector<int const>{};     // error
    auto _ = std::vector<int volatile>{};  // error
    auto _ = std::vector<int&>{};          // error
}

See also https://godbolt.org/z/3GY9E66xh

This is because std::allocator<T> doesn't accept T if it is qualified by const, volatile, or reference. I just wonder:

Is it required by the C++ standard that T must not be qualified by const, volatile, or reference in std::allocator<T>?

have you read the error message you get when doing one of these? — Marcus Müller
– Marcus Müller, Commented Jul 11 at 9:37
static_assert(std::is_object_v<Key>, "The C++ Standard forbids containers of non-object types because of [container.requirements]."); — 許恩嘉
– 許恩嘉, Commented Jul 11 at 9:39
@MarcusMüller libc++ uses static_assert to ensure value_type is unqualified. I just wonder if that is also explicitly required by the C++ standard. — xmllmx
– xmllmx, Commented Jul 11 at 9:40
en.cppreference.com/w/cpp/named_req/Allocator.html -- I think that should answer it. — Ulrich Eckhardt
– Ulrich Eckhardt, Commented Jul 11 at 10:04

xmllmx · Accepted Answer · 2025-07-11 11:18:49Z

8

Excerpts from the lastest C++ standard n5008 (emphasis mine):

Section 16.4.4.6.1/p2 [allocator.requirements.general]

T, U, C denote any cv-unqualified object type,

X denotes an allocator class for type T,

Section 6.8.1/p8 [allocator.requirements.general]

An object type is a (possibly cv-qualified) type that is not a function type, not a reference type, and not cv void.

For std::allocator<T>, T must be cv-unqualified object type, and an object type must not be a reference type. So we can say:

The C++ standard explicitly requires the template parameter type T in std::allocator<T> must not be qualified by const, volatile, or reference.

edited Jul 11 at 11:18

answered Jul 11 at 10:54

xmllmx

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