how do I print an unsigned char as hex in c++ using ostream?

I want to work with unsigned 8-bit variables in C++. Either unsigned char or uint8_t do the trick as far as the arithmetic is concerned (which is expected, since AFAIK uint8_t is just an alias for unsigned char, or so the debugger presents it.

The problem is that if I print out the variables using ostream in C++ it treats it as char. If I have:

unsigned char a = 0;
unsigned char b = 0xff;
cout << "a is " << hex << a <<"; b is " << hex << b << endl;

then the output is:

a is ^@; b is 377

instead of

a is 0; b is ff

I tried using uint8_t, but as I mentioned before, that's typedef'ed to unsigned char, so it does the same. How can I print my variables correctly?

Edit: I do this in many places throughout my code. Is there any way I can do this without casting to int each time I want to print?


Use:

cout << "a is " << hex << (int) a <<"; b is " << hex << (int) b << endl;

And if you want padding with leading zeros then:

#include <iomanip>
...
cout << "a is " << setw(2) << setfill('0') << hex << (int) a ; 

As we are using C-style casts, why not go the whole hog with terminal C++ badness and use a macro!

#define HEX( x )
   setw(2) << setfill('0') << hex << (int)( x )

you can then say

cout << "a is " << HEX( a );

Edit: Having said that, MartinStettner's solution is much nicer!


I would suggest using the following technique:

struct HexCharStruct
{
  unsigned char c;
  HexCharStruct(unsigned char _c) : c(_c) { }
};

inline std::ostream& operator<<(std::ostream& o, const HexCharStruct& hs)
{
  return (o << std::hex << (int)hs.c);
}

inline HexCharStruct hex(unsigned char _c)
{
  return HexCharStruct(_c);
}

int main()
{
  char a = 131;
  std::cout << hex(a) << std::endl;
}

It's short to write, has the same efficiency as the original solution and it lets you choose to use the "original" character output. And it's type-safe (not using "evil" macros :-))


You can read more about this at http://cpp.indi.frih.net/blog/2014/09/tippet-printing-numeric-values-for-chars-and-uint8_t/ and http://cpp.indi.frih.net/blog/2014/08/code-critique-stack-overflow-posters-cant-print-the-numeric-value-of-a-char/. I am only posting this because it has become clear that the author of the above articles does not intend to.

The simplest and most correct technique to do print a char as hex is

unsigned char a = 0;
unsigned char b = 0xff;
auto flags = cout.flags(); //I only include resetting the ioflags because so
                           //many answers on this page call functions where
                           //flags are changed and leave no way to  
                           //return them to the state they were in before 
                           //the function call
cout << "a is " << hex << +a <<"; b is " << +b << endl;
cout.flags(flags);

The readers digest version of how this works is that the unary + operator forces a no op type conversion to an int with the correct signedness. So, an unsigned char converts to unsigned int, a signed char converts to int, and a char converts to either unsigned int or int depending on whether char is signed or unsigned on your platform (it comes as a shock to many that char is special and not specified as either signed or unsigned).

The only negative of this technique is that it may not be obvious what is happening to a someone that is unfamiliar with it. However, I think that it is better to use the technique that is correct and teach others about it rather than doing something that is incorrect but more immediately clear.