Why do I have to cast a byte as unsigned twice to see hex output?

Question

I want to print a variable as hex:

#include <iostream>
#include <string>
#include <cstdint>

int main() {
    auto c = 0xb7;

    std::cout << std::hex << static_cast<unsigned char>(c) << std::endl;
    std::cout << std::hex << static_cast<unsigned>(static_cast<unsigned char>(c)) << std::endl;

    std::cout << std::hex << (uint8_t)(c) << std::endl;
    std::cout << std::hex << (unsigned)(uint8_t)(c) << std::endl;

    return 0;
}

The output seems to be:

\ufffd (tries to print it as a char)
b7
\ufffd (tries to print it as a char)
b7

I do understand that c has higher bits set (10110111), but I cast it to uint8_t and unsigned char once already.

Why do I have to cast uint8_t or unsigned char to unsigned again to get the expected output?

Perhaps would be better not to use `auto` in this case - be clear on the type — Ed Heal, Dec 11 '15 at 19:57
@5gon12eder, Ed: This question *is* about understanding types. That's why I explicitly used auto (what does the compiler guess it as?) and did not print out not using casting — PoorLuzer, Dec 11 '15 at 19:59
@PoorLuzer The deduced type of `c` is `int`. Maybe you should change your title. It sounds a lot like you want to get hex output working. — 5gon12eder, Dec 11 '15 at 20:01
@PoorLuzer I would but I'm not sure what question you want to ask. — 5gon12eder, Dec 11 '15 at 20:03
Wow, all those calls to `endl`! `std::cout` will most certainly be flushed! (use '\n' when all you need is a new line) — Pete Becker, Dec 11 '15 at 20:16
You assuredly don't need to cast twice. I note that you never tried casting directly to `unsigned` without going through `unsigned char`. It seems that your real question is "Why does sending an `unsigned char` to `std::cout` print it as a character?" Are you really surprised by that? Which leaves only "Why is `uint8_t` treated as a character?", which is already answered, so I've marked a duplicate. — Ben Voigt, Dec 11 '15 at 20:22
_"Why do I have to cast a byte as unsigned twice to see hex output?"_ You don't, and you didn't. — Lightness Races in Orbit, Dec 11 '15 at 20:30
@BenVoigt: Are you implying the `uint8_t` is treated as `unsigned char` ? — PoorLuzer, Dec 11 '15 at 22:29

YSC · Accepted Answer · 2015-12-11T20:28:05.103

1

std::hex sets the basefield of the stream str to hex as if by calling str.setf(std::ios_base::hex, std::ios_base::basefield).

When this basefield hex bit is set, iostreams use hexadecimal base for integer I/O.

Code

#include <iostream>

int main()
{
    int i = 0xb7;
    unsigned u = 0xb7;
    char c = static_cast<char>(0xb7);
    unsigned char b = 0xb7;

    std::cout << std::hex << i << std::endl;
    std::cout << std::hex << u << std::endl;
    std::cout << std::hex << c << std::endl;
    std::cout << std::hex << b << std::endl;
    return 0;
}

Output

b7
b7
�
�

I suspect this output to vary on a Windows (non UTF-8) system.

edited Dec 11 '15 at 20:28

answered Dec 11 '15 at 20:19

YSC

38,212
9
96
149

So does this mean that unless the type is an `int`, `iostreams` will ignore the `std::hex` modifier and print everything as a raw byte anyways? Even an `uint8_t` which I assumed to be a byte (incorrect assumption?) – PoorLuzer Dec 11 '15 at 23:48
Doesn't have to be `int` in particular, does have to be an integral type that is not a character type. Standard doesn't say if `uint8_t` is a character type - it is allowed to be but not required. On your system it is. See linked question. – Ben Voigt Dec 12 '15 at 00:24
Great answer. Loved the references. I'll email you shortly at the contact info. Would you be able to review some of my unanswered questions? (two of them below) – PoorLuzer Dec 28 '15 at 22:29
http://stackoverflow.com/questions/873356/challenges-in-writing-wrappers-for-c-functions-so-that-they-can-be-used-from-c http://stackoverflow.com/questions/874161/c-tokenizer-and-it-returns-empty-too-when-fields-are-missing-yay – PoorLuzer Dec 28 '15 at 22:29

Why do I have to cast a byte as unsigned twice to see hex output?

1 Answers1

Code

Output