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

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How do I print a hexadecimal character?

Here's a solution: char ch = 0xC0; printf("%x", ch & 0xff);

How do I print unsigned characters?

unsigned char ch = 212 ; And your printf will work. Even with ch changed to unsigned char , the behavior of the code is not defined by the C standard. This is because the unsigned char is promoted to an int (in normal C implementations), so an int is passed to printf for the specifier %u .

How to print char as hex in C++?

Method One: std::cout Method 1 as shown above is probably the more C++ way: Cast to an unsigned int. Use std::hex to represent the value as hexadecimal digits. Use std::setw and std::setfill from <iomanip> to format.

What is an unsigned char?

unsigned char is a character datatype where the variable consumes all the 8 bits of the memory and there is no sign bit (which is there in signed char). So it means that the range of unsigned char data type ranges from 0 to 255.


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 :-))


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!


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.


Well, this works for me:

std::cout << std::hex << (0xFF & a) << std::endl;

If you just cast (int) as suggested it might add 1s to the left of a if its most significant bit is 1. So making this binary AND operation guarantees the output will have the left bits filled by 0s and also converts it to unsigned int forcing cout to print it as hex.

I hope this helps.


In C++20 you'll be able to use std::format to do this:

std::cout << std::format("a is {:x}; b is {:x}\n", a, b);

Output:

a is 0; b is ff

In the meantime you can use the {fmt} library, std::format is based on. {fmt} also provides the print function that makes this even easier and more efficient (godbolt):

fmt::print("a is {:x}; b is {:x}\n", a, b);

Disclaimer: I'm the author of {fmt} and C++20 std::format.


Hm, it seems I re-invented the wheel yesterday... But hey, at least it's a generic wheel this time :) chars are printed with two hex digits, shorts with 4 hex digits and so on.

template<typename T>
struct hex_t
{
    T x;
};

template<typename T>
hex_t<T> hex(T x)
{
    hex_t<T> h = {x};
    return h;
}

template<typename T>
std::ostream& operator<<(std::ostream& os, hex_t<T> h)
{
    char buffer[2 * sizeof(T)];
    for (auto i = sizeof buffer; i--; )
    {
        buffer[i] = "0123456789ABCDEF"[h.x & 15];
        h.x >>= 4;
    }
    os.write(buffer, sizeof buffer);
    return os;
}