if-constexpr

Maybe I missed something, but I can't find any hints: is there a constexpr ternary operator in C++17 equivalent to constexpr-if? template<typename Mode> class BusAddress { public: explicit constexpr BusAddress(Address device) : mAddress(Mode::write ? (device.mDevice << 1) : (device.mDevice << 1) | 0x01) {} private: uint8_t mAddress = 0; }; 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

I have compiled and ran the following program in a C++17 compiler (Coliru). In the program, I declared an extern variable, but did not define it. However, the compiler doesn't give a linker error . #include <iostream> extern int i; // Only declaration int func() { if constexpr (true) return 0; else if (i) return i; else return -1; } int main() { int ret = func(); std::cout<<"Ret : "<<ret<<std::endl; } Why doesn't the compiler give a linker error? Because the variable isn't odr-used. You have a constexpr if there that always discards the branch that could use it. One of the points of constexpr

I'm trying to compare a function parameter inside a constexpr-if statement. Here is a simple example: constexpr bool test_int(const int i) { if constexpr(i == 5) { return true; } else { return false; } } However, when I compile this with GCC 7 with the following flags: g++-7 -std=c++1z test.cpp -o test I get the following error message: test.cpp: In function 'constexpr bool test_int(int)': test.cpp:3:21: error: 'i' is not a constant expression if constexpr(i == 5) { return true; } However, if I replace test_int with a different function: constexpr bool test_int_no_if(const int i) { return (i =

I want to use a constexpr bool ( useF in the example below) to enable a feature in the following code. Here, calling A::f() . Additionally, I want to be the alias-template ( a ) to be void in the case I switch off the feature. I tried to use a constexpr if statement, but the body is still being instantiated, which causes a compile error. If I use a wrapper template ( X ), the body is being discarded as I'd expected, but that seems ugly to me. Are there any other ways to do this? constexpr bool useF = false; struct A { static void f() {} }; using a = std::conditional<useF, A, void>::type;

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 check should be left unpreformed on a discarded branch. Making an indirection via a constexpr lambda does

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 (condition) { return 0; } else { // optimized out even without "constexpr if" return 1; } } Godbolt

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 " << value << std::endl; } int main() { auto n = 1000; print(n); print(&n); } But when I rewrite the above