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Home/ Questions/Q 1060621
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Editorial Team
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Editorial Team
Asked: May 16, 20262026-05-16T18:21:18+00:00 2026-05-16T18:21:18+00:00

I have this code: string get_md5sum(unsigned char* md) { char buf[MD5_DIGEST_LENGTH + MD5_DIGEST_LENGTH]; char

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I have this code:

    string get_md5sum(unsigned char* md) {
        char buf[MD5_DIGEST_LENGTH + MD5_DIGEST_LENGTH];
        char *bptr;
        bptr = buf;
        for(int i = 0; i < MD5_DIGEST_LENGTH; i++) {
                bptr += sprintf(bptr, "%02x", md[i]);
        }
        bptr += '\0';
        string x(buf);
        return x;
}

Unfortunately, this is some C combined with some C++. It does compile, but I don’t like the printf and char*’s. I always thought this was not necessary in C++, and that there were other functions and classes to realize this. However, I don’t completely understand what is going on with this:

 bptr += sprintf(bptr, "%02x", md[i]);

And therefore I don’t know how to convert it into C++. Can someone help me out with that?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-16T18:21:18+00:00Added an answer on May 16, 2026 at 6:21 pm

    sprintf returns number of bytes written. So this one writes to bptr two bytes (value of md[i] converted to %02x -> which means hex, padded on 2 chars with zeroes from left), and increases bptr by number of bytes written, so it points on string’s (buf) end.

    I don’t get the bptr += '\0'; line, IMO it should be *bptr = '\0';

    in C++ it should be written like this:

    using namespace std;
    stringstream buf;
    for(int i = 0; i < MD5_DIGEST_LENGTH; i++)
    {
        buf << hex << setfill('0') << setw(2) << static_cast<int>(static_cast<unsigned char>(md[i])); 
    }
    return buf.str();
    

    EDIT: updated my c++ answer

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