C++：将字节数组序列化为十六进制字符串

Question

I (C++ newbie) am currently trying to implement the following function:

std::string bytes_to_hex(const std::string &bytes);

The function should basically return a base16 encoding of a given byte array:

std::string input{0xde, 0xad, 0xbe, 0xef} => "deadbeef"

My first version doesn't quite work how I imagined:

std::string bytes_to_hex(const std::string &bytes) {
    std::ostringstream ss;
    ss << std::hex;

    for (auto &c : bytes) {
        ss << std::setfill('0') << std::setw(2) << +c;
    }

    return ss.str();
}

With this function the output is:

ffffffdeffffffadffffffbeffffffef

After some experiments, I've found out that this version looks better:

std::string bytes_to_hex(const std::string &bytes) {
    std::ostringstream ss;
    ss << std::hex;

    for (const char &c : bytes) {
        ss << std::setfill('0') << std::setw(2) << +(static_cast<uint8_t>(c));
    }

    return ss.str();
}

The output is as expected:

deadbeef

My question is:

• Why does the second version work and the first doesn't? What is the main difference here?
• Is the second version correct implementation of my original intention or can there be other problems?

Answer 1

As mentioned in my comment, the unary + forces integer promotion . When that happens, signed types are sign extened which for two's complement encoded integers means that negative numbers (where the left-most bit is 1 ) are left-padded with binary ones (ie 0xde becomes 0xffffffde ).

Also mentioned is that char can be either signed or unsigned , a decision that is up to the compiler. Because of the output you get we can say that in your case char is actually signed char .

The simple solution you found out is to first cast the character to an unsigned char , and then (with the unary + ) promote it to int .