You can't static_cast in this case; https://godbolt.org/z/a1bMbPcaj

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

No, you can't do that either: 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

Good catch. Its weird that it compiles without error as a consteval func.

Hmm, 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.

Ah 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!