if-constexpr

问题 In reference to this question. The core constant expression that is used to initialize the constexpr variable y is ill-formed. So much is a given. But if I try to turn the if into an if constexpr : template <typename T> void foo() { constexpr int x = -1; if constexpr (x >= 0){ constexpr int y = 1 << x; } } int main(){ foo<int>(); } The error persists. With GCC 7.2 still giving: error: right operand of shift expression '(1 << -1)' is negative [-fpermissive] But I thought that the semantic

问题 C++1z will introduce "constexpr if" - an if that will have one of branches removed, based on the condition. Seems reasonable and useful. However, is it not possible to do without constexpr keyword? I think that during compilation, compiler should know wheter condition is known during compilation time or not. If it is, even the most basic optimization level should remove unnecessary branch. For example (see in godbolt: https://godbolt.org/g/IpY5y5): int test() { const bool condition = true; if

问题 I tried to play with the C++17 standard. I tried to use one of the features of C++17 if constexpr . And I had a problem... Please take a look at the following code. This compiles without errors. In the following code, I tried to use if constexpr to check if it is a pointer. #include <iostream> #include <type_traits> template <typename T> void print(T value) { if constexpr (std::is_pointer_v<decltype(value)>) std::cout << \"Ptr to \" << *value << std::endl; // Ok else std::cout << \"Ref to \"