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>?

Answer 1

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.