The following code comes from this StackOverflow Q&A (note this is not related to std::bitset, it's just a homegrown class to experiment with constexpr): http://stackoverflow.com/q/29871138/819272

#include <cstddef>

template<std::size_t>
class bitset;

template<std::size_t N>
constexpr bool operator==(const bitset<N>&, const bitset<N>&) noexcept;

template<std::size_t N>
class bitset
{
    friend constexpr bool operator== <>(const bitset<N>&, const bitset<N>&) noexcept;
    //^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ <-- error from this piece
};

template<std::size_t N>
constexpr bool operator==(const bitset<N>&, const bitset<N>&) noexcept
{
    return true;
}

int main() {}

The error I get with -std=c++11 on gcc 4.9.1 through gcc HEAD 6.0.0 20150501 is

'constexpr' is not allowed in declaration of friend template specialization 

The same code on older gcc version (4.6 through 4.8) gives a slightly different error

'inline' is not allowed in declaration of friend template specialization

which seems to have been introduced in 2013 https://gcc.gnu.org/ml/gcc-patches/2013-03/msg01028.html

Richard Smith's accepted answer on StackOverflow ( http://stackoverflow.com/a/29957648/819272 ) states that although constexpr functions are implicitly inline, an inline function does not imply the inline specifier. Therefore the above code should be valid.