3 ms·
You can use `std::bit_cast` to do that in constexpr contexts. constexpr auto f(uint8_t *x) { return std::bit_cast<char *>(x); } https://godbolt.
by lbhdc 1y ago
You can use `std::bit_cast` to do that in constexpr contexts.
constexpr auto f(uint8_t *x) {
return std::bit_cast<char *>(x);
}
https://godbolt.org/z/K3f9b9GGs https://godbolt.org/z/K3f9b9GGs
- psyclobe 1y agoAh this was my case! Was trying to constexpr a uint8_t ptr to char * in a constexpr constructor for a string class. Ah that’s what bitcast is for, neat!
- TuxSH 1y agoNo, you can't do that either: https://godbolt.org/z/vzdTMazx7 https://godbolt.org/z/vzdTMazx7 : error: '__builtin_bit_cast' is not a constant expression because 'char' is a pointer type Here the `constexpr` keyword means the function might be called in a constant-evaluated context. f doesn't need to have all its statements be able to be evaluated in constexpr, only those which are actually used are. You need to explicitly instantiate a constexpr variable to test this. cppreference is very clear* about this, regarding bit_cast: https://en.cppreference.com/w/cpp/numeric/bit_cast https://en.cppreference.com/w/cpp/numeric/bit_cast
- lbhdc 1y agoGood catch. Its weird that it compiles without error as a consteval func.
- TuxSH 1y agoHmm, looking at cppreference: The consteval specifier declares a function or function template to be an immediate function, that is, every potentially-evaluated call to the function must (directly or indirectly) produce a compile time constant expression. It's possible that the compiler just doesn't bother as long as you aren't actually calling the function.