How can I print 0x0a instead of 0xa using cout?

Question

How can I print 0x0a, instead of 0xa using cout?

#include  <iostream>  using std::cout;   using std::endl;   using std::hex;  int main()   {       cout << hex << showbase << 10 << endl;   } 

Answer 1

This works for me in GCC:

#include  <iostream> #include  <iomanip>  using namespace std;  int main() {     cout << "0x" << setfill('0') << setw(2) << right << hex << 10 << endl; } 

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If you are getting sick and tired of iostream's formatting quirkiness, give Boost.Format a try. It allows good-old-fashioned, printf-style format specifiers, yet it is type-safe.

#include <iostream> #include <boost/format.hpp>  int main() {     std::cout << boost::format("0x%02x\n") % 10; } 

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UPDATE (2019)

Check out the {fmt} library that's been accepted into C++20. Benchmarks show it to be faster than Boost.Format.

#if __has_include(<format>)     #include <format>     using std::format; #else     #include <fmt/format.h>     using fmt::format; #endif  std::cout << format("{:#04x}\n", 10); 

Answer 2

Use setw and setfill from iomanip

#include  <iostream> #include  <iomanip>  using std::cout;   using std::endl;   using std::hex;  int main() {     cout << "0x" << std::setfill('0') << std::setw(2) << hex << 10 << endl; } 

Personally, the stateful nature of iostreams always annoys me. I think boost format is a better option, so I'd recommended the other answer.