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lbhdc • 04/24/2025 • 2 replies • view on HN

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

Replies

TuxSH • 04/24/2025

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

➕ show 1 reply
psyclobe • 04/24/2025

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!