How to create an octet type similar to std::byte in C++?

Question

The type std::byte comprises CHAR_BIT bits, and that may be more than 8.

So, how do I declare a true octet in C++?

Answer 1

If the hardware supports 8-bit integers with no padding, then std::uint8_t will be defined, and could be used for that purpose.

enum class octet : std::uint8_t {};

// add bitwise ops similar to the operator overloads for std::byte
// alternatively, use std::uint8_t directly

However, this is basically pointless. If unsigned char is wider than 8 bits, then std::uint8_t won't be defined because no type can be smaller than a byte in C++.

Any "true octet" type is either not going to be portable, or it will require padding. If padding is acceptable, you can define it as follows:

enum class octet : std::uint_least8_t {};

constexpr bool operator==(octet x, octet y) noexcept {
    return (unsigned(x) & 0xff) == (unsigned(y) & 0xff);
}

// TODO: implement other comparisons

constexpr octet operator|(octet x, octet y) noexcept {
    return octet(unsigned(x) | unsigned(y));
}

constexpr octet operator<<(octet x, std::integral auto s) noexcept {
    return octet((unsigned(x) << s) & 0xff);
}

// TODO: implement other bitwise ops, and possibly arithmetic ops

Also, as commenters have pointed out, hardware with no support for 8-bit types is extremely rare. Unless you expect to compile your code for digital signal processors (DSPs) or other unusual hardware, just assume 8-bit bytes. Use std::uint8_t, or

static_assert(CHAR_BIT == 8);