how do I declare and initialize an array of bytes in C++

Question

Is this really the best way to declare a byte (or array of bytes)?

static constexpr byte kGuard1_[] = {
            byte{0x45}, byte{0x23}, byte{0x12}, byte{0x56}, byte{0x99}, byte{0x76}, byte{0x12}, byte{0x55},
        };

why isn't there some suffix (like b) that you can use to directly mark the number as a byte? Or is the problem just with my use of uniform initialization?

Answer 1

I can't say if this is better but it's different since you won't have to repeat byte for every element:

#include <array>
#include <utility>

template <class T, class... Args>
constexpr auto mkarr(Args&&... args) {
    return std::array{static_cast<T>(std::forward<Args>(args))...};
}

static constexpr auto kGuard1 = mkarr<std::byte>(0x45, 0x23, 0x12, 0x56,
                                                 0x99, 0x76, 0x12, 0x55);

Note that it uses a std::array<std::byte, 8> instead of a std::byte[8].

---

why isn't there some suffix (like b) that you can use to directly mark the number as a byte?

I can't say, but if you want you can define your own user-defined literal that could be used with C arrays.

constexpr std::byte operator""_B(unsigned long long x) {
    // note: no space between "" and _B above
    return static_cast<std::byte>(x); 
}

static constexpr std::byte kGuard1[]{0x45_B, 0x23_B, 0x12_B, 0x56_B,
                                     0x99_B, 0x76_B, 0x12_B, 0x55_B};

Or is the problem just with my use of uniform initialization?

No, it's just that there is no implicit conversion to std::byte.