std::sqrt is not defined as constexpr, according to section 26.8 of N3291: the C++11 FDIS (and I doubt they added it to the final standard after that). One could possibly write such a version, but the standard library version is not constexpr.
Think about what constexpr means. It means that I can resolve this value at compile time. Thus, a member variable of a class cannot itself be a constexpr…the instance that xVal belongs to does not exist until instantiation time! The thing that owns xVal could be constexp, and that would make xVal a constexpr, but … Read more
constexpr variables are not compile-time values A value is immutable and does not occupy storage (it has no address), however objects declared as constexpr can be mutable and do occupy storage (under the as-if rule). Mutability Most objects declared as constexpr are immutable, but it is possible to define a constexpr object that is (partially) … Read more
I agree that this feature is missing – I need it too. Example: double pow(double x, int n) { // calculate x to the power of n return … } static inline double pow (double x, constexpr int n) { // a faster implementation is possible when n is a compile time constant return … … Read more
I don’t think this is a bug. If you change the constexpr to const, it still fails, with the exact same error. You’ve declared S::X, but not defined it anywhere, so there’s no storage for it. If you do anything with it that needs to know the address of it then you’ll need to define … Read more
constexpr functions Functions have an advantage that free variables do not have (until C++14 that is): they can easily be templated without some class boilerplate. That means you can have your pi with a precision depending on a template argument: template<typename T> constexpr T pi(); template<> constexpr float pi() { return 3.14f; } template<> constexpr … Read more
This is not valid C++11 code, because a constexpr function must only contain a return statement. This is incorrect. static_assert in a constexpr function are fine. What is not fine is using function parameters in constant expressions, like you do it. You could throw if x <= 0. Calling the function in a context that … Read more
This is talking about a well-established rule for templates – the same rule that allows compilers to diagnose template<class> void f() { return 1; }. [temp.res]/8 with the new change bolded: The program is ill-formed, no diagnostic required, if: no valid specialization can be generated for a template or a substatement of a constexpr if … Read more
Template integral_constant defines a type, keyword constexpr defines a constant. For example std::true_type is std::integral_constant<bool, true>. One of the usage examples is tag-dispatching. template<typename T> void use_impl(const T&, std::false_type) { } template<typename T> void use_impl(const T&, std::true_type) { } template<typename T> void use(const T& v) { use_impl(v, typename std::is_integral<T>::type()); } Live example
No, there is no constexepr conditional operator. But you could wrap the whole thing in a lambda and immediately evaluate it (an IIFE): template<typename Mode> class BusAddress { public: explicit constexpr BusAddress(Address device) : mAddress([&]{ if constexpr (Mode::write) { return device.mDevice << 1; } else { return (device.mDevice << 1) | 0x01; } }()) { … Read more