This is just how constexpr if works. If we check [stmt.if]/2 If the if statement is of the form if constexpr, the value of the condition shall be a contextually converted constant expression of type bool; this form is called a constexpr if statement. If the value of the converted condition is false, the first … Read more
This is just how constexpr if works. If we check [stmt.if]/2 If the if statement is of the form if constexpr, the value of the condition shall be a contextually converted constant expression of type bool; this form is called a constexpr if statement. If the value of the converted condition is false, the first … Read more
Here’s another solution, which is more generic (applicable to any expression, without defining a separate template each time). This solution leverages that (1) lambda expressions can be constexpr as of C++17 (2) the type of a captureless lambda is default constructible as of C++20. The idea is, the overload that returns true is selected when … Read more
Do we need to put constexpr after every if statement in if-else block in these kind of situations? Yes. The else-if block1 is a lie :), there are only if blocks1 and else blocks1. This is how your code is seen by the compiler: if constexpr (std::is_same_v<int, T>) return {a, 0.0}; else // { if … Read more
This is easy to explain through an example. Consider struct Cat { void meow() { } }; struct Dog { void bark() { } }; and template <typename T> void pet(T x) { if(std::is_same<T, Cat>{}){ x.meow(); } else if(std::is_same<T, Dog>{}){ x.bark(); } } Invoking pet(Cat{}); pet(Dog{}); will trigger a compilation error (wandbox example), because both … Read more
The standard doesn’t say much about the discarded statement of an if constexpr. There are essentially two statements in [stmt.if] about these: In an enclosing template discarded statements are not instantiated. Names referenced from a discarded statement are not required ODR to be defined. Neither of these applies to your use: the compilers are correct … Read more
I would like to know why “if constexpr” works only in template functions, even if the type is deduced by the decltype from the input parameter. This is by design. if constexpr will not instantiate the branch not taken if it’s within a template. It won’t just treat the branch not taken as token soup … Read more