Converting hexadecimal to uint8_t in C++

Question

I am working in C++ and let's say I have the following hexadecimal as a string

string key = "F3D5";

how can I convert it to an array of uint8_t?

Actually, I have an algorithm that generate a key of type uint8_t*, and I convert this key to hexadecimal using the following code :

uint8_t key = generateKey();
vector<uint8_t> keyVector(begin(key), end(key));
string hexadecimalKey= uint8_vector_to_hex_string(keyVector);

This is the method that convert uint8 vector to hexadecimal in string :

string uint8_vector_to_hex_string(const vector<uint8_t>& v) {
    stringstream ss;
    ss << hex << setfill('0');
    vector<uint8_t>::const_iterator it;

    for (it = v.begin(); it != v.end(); it++) {
        ss << setw(2) << static_cast<unsigned>(*it);
    }

    return ss.str();
}

And I would like to re-convert the hexadecimalKey to uint8_t, with the following code

uint8_t* convertedKey = hex_string_to_uint8_t(hexadecimalKey);

with the following method :

uint8_t* hex_string_to_uint8_t(string value) {
    size_t len = value.length();
    vector<uint8_t> out;
    for (size_t i = 0; i < len; i += 2) {
        std::istringstream strm(value.substr(i, 2));
        uint8_t x;
        strm >> std::hex >> x;
        out.push_back(x);
    }
    return out.data();
}

and whenever i do that, the convertedKey is not equal to key

I need your help.

Thank you, really appreciate it

Possible duplicate of [atoi() from hex representation string](https://stackoverflow.com/questions/20654424/atoi-from-hex-representation-string) — Axalo, Mar 24 '18 at 19:38
You shoud consider adding some details and at the very least what kind of array (using the STL, C-style array, etc.). — Paradox, Mar 24 '18 at 19:39
You can’t return the `data()` of a local `vector`: it’s destroyed when the container is. — Davis Herring, Mar 24 '18 at 22:15

score 4 · Accepted Answer · edited Mar 26 '18 at 10:21

You can use the function below:

uint8_t* hex_str_to_uint8(const char* string) {

    if (string == NULL)
        return NULL;

    size_t slength = strlen(string);
    if ((slength % 2) != 0) // must be even
        return NULL;

    size_t dlength = slength / 2;

    uint8_t* data = (uint8_t*)malloc(dlength);

    memset(data, 0, dlength);

    size_t index = 0;
    while (index < slength) {
        char c = string[index];
        int value = 0;
        if (c >= '0' && c <= '9')
            value = (c - '0');
        else if (c >= 'A' && c <= 'F')
            value = (10 + (c - 'A'));
        else if (c >= 'a' && c <= 'f')
            value = (10 + (c - 'a'));
        else
            return NULL;

        data[(index / 2)] += value << (((index + 1) % 2) * 4);

        index++;
    }

    return data;
}

Good luck!

This is not good C++ code. Using `malloc` is a bit worse than `new[]` and quite a bit worse than `std::vector`. I know the question specified `uint8_t*` as the return type, but this is clearly a question from a C++ novice and you should not repeat the mistakes of the question. — MSalters, Mar 26 '18 at 11:10

score -1 · Answer 2 · answered Mar 24 '18 at 22:24

List item

Looks like it works, but you should work on lining everything up -- make everything in the same block (section of code with {}) line up, and then intent after a { and put each } on a new line. Or follow some other coding standard, I don't care. Just make it line up and indent.

Also, you could make it ever so slightly more efficient by making tmp just a char (not an array) and then using it to store the third character, setting that character to NULL, and then restoring it after strtol.

char buf[] = "4142434445";
uint8_t tx_buffer[20];
uint8_t len_buffer=0;


for(i=0;i<strlen(buf);i+=2) {
char tmp = buf[i+3];
buf[i+3] = 0;
tx_buffer[len_buffer] = strtol(buf,NULL,16);
len_buffer++;
buf[i+3] = tmp;
}

Converting hexadecimal to uint8_t in C++

2 Answers2