On Sunday 01 September 2013 11:06:16 Daryle Walker wrote:
I don’t think you need a “constexpr_if.” You can unconditionally put BOOST_CONSTEXPR in the definition. This will unconditionally add “constexpr” to the member operator (for C++11 systems), which will be ignored if the compiler determines that the attached function (or a particular call) is disqualified from being used in constant expressions. So any extra preprocessor work, or a variant macro, is unnecessary.
No, that doesn't work. struct foo { void* m_p; constexpr explicit operator bool () const { return !this->operator!(); } bool operator! () const { return !m_p; } }; g++ -std=gnu++11 constexpr_operator.cpp -o constexpr_operator constexpr_operator.cpp: In member function ‘constexpr foo::operator bool() const’: constexpr_operator.cpp:9:27: error: call to non-constexpr function ‘bool foo::operator!() const’