Char array to hex string conversion - unexpected output

Question

I have a simple program converting dynamic char array to hex string representation.

#include <iostream>
#include <sstream>
#include <iomanip>
#include <string>
using namespace std;


int main(int argc, char const* argv[]) {
    int length = 2;
    char *buf = new char[length];
    buf[0] = 0xFC;
    buf[1] = 0x01;

    stringstream ss;
    ss << hex << setfill('0');
    for(int i = 0; i < length; ++i) {
        ss << std::hex << std::setfill('0') << std::setw(2) << (int) buf[i] << " ";
    }
    string mystr = ss.str();
    cout << mystr << endl;
}

Output:

fffffffc 01

Expected output:

fc 01

Why is this happening? What are those ffffff before fc? This happens only on certain bytes, as you can see the 0x01 is formatted correctly.

Answer 1

Three things you need to know to understand what's happening:

1. The first thing is that char can be either signed or unsigned, it's implementation (compiler) specific

2. When converting a small signed type to a large signed type (like e.g. a signed char to an int), they will be sign extended

3. How negative values are stored using the most common two's complement system, where the highest bit in a value defines if a value is negative (bit is set) or not (bit is clear)

What happens here is that char seems to be signed, and 0xfc is considered a negative value, and when you convert 0xfc to an int it will be sign-extended to 0xfffffffc.

To solve it use explicitly unsigned char and convert to unsigned int.