Printing zero-padded hex with std::cout

Question

Say I have a dword I want to output in hex with std::cout and left-pad with zeros, so 0xabcd will be shown as 0x0000abcd. It seems like you would have to do this:

uint32_t my_int = 0xabcd;

std::cout << "0x" << std::hex << std::setw(8) << std::setfill('0')
    << my_int << std::endl;

This seems ridiculous for something that can be accomplished in C with printf("0x%08X\n", my_int);. Is there any way to make this shorter while still using std::cout for output (besides using namespace std)?

Answer 1

I suppose you can write a "stream manipulator". This is useful if you have multiple hex numbers you want to print in this format. This is clearly not an ideal solution, but using a wrapper type you can make your own "format flag" to toggle it. See Sticky custom stream manipulator for more information.

#include <iostream>
#include <iomanip>

static int const index = std::ios_base::xalloc();

std::ostream& hexify(std::ostream& stream) {
    stream.iword(index) = 1;
    return stream;
}

std::ostream& nohexify(std::ostream& stream) {
    stream.iword(index) = 0;
    return stream;
}

struct WrapperType {
    uint32_t _m;
public:
    WrapperType(uint32_t m) : _m(m)
    {
    }

    uint32_t getm() const
    {
        return _m;
    }
};
std::ostream& operator<< (std::ostream& os, const WrapperType& t) {
    if (os.iword(index))
        return os << "0x" << std::hex << std::setw(8) << std::setfill('0') << t.getm();
    else
        return os << t.getm();
}

int main()
{
    WrapperType my_int{0xabcd};
    std::cout << hexify << my_int << my_int;
    std::cout << nohexify << my_int;
}