uint8_t iostream behavior

Question

Abstract: I was expecting the code: cout << uint8_t(0); to print "0", but it doesn't print anything.

Long version: When I try to stream uint8_t objects to cout, I get strange characters with gcc. Is this expected behavior? Could it be that uint8_t is an alias for some char-based type? See compiler/system notes in the code example.

// compile and run with:
// g++ test-uint8.cpp -std=c++11 && ./a.out
//                    -std=c++0x (for older gcc versions)
/**
 * prints out the following with compiler:
 *     gcc (GCC) 4.7.2 20120921 (Red Hat 4.7.2-2)
 * on the system:
 *     Linux 3.7.9-101.fc17.x86_64
 * Note that the first print statement uses an unset uint8_t
 * and therefore the behaviour is undefined. (Included here for
 * completeness)

> g++ test-uint8.cpp -std=c++11 && ./a.out
>>>�<<<    >>>194<<<
>>><<<    >>>0<<<
>>><<<    >>>0<<<
>>><<<    >>>0<<<
>>><<<    >>>1<<<
>>><<<    >>>2<<<

 *
 **/

#include <cstdint>
#include <iostream>

void print(const uint8_t& n)
{
    std::cout << ">>>" << n                 << "<<<    "
              << ">>>" << (unsigned int)(n) << "<<<\n";
}

int main()
{
    uint8_t a;
    uint8_t b(0);
    uint8_t c = 0;
    uint8_t d{0};
    uint8_t e = 1;
    uint8_t f = 2;
    for (auto i : {a,b,c,d,e,f})
    {
        print(i);
    }
}

Answer 1

uint8_t is an alias for unsigned char, and the iostreams have special overloads for chars that print out the characters rather than formatting numbers.

The conversion to integer inhibits this.

Answer 2

Could it be that uint8_t is an alias for some char-based type?

Absolutely. It's required to be a typedef for a builtin 8-bit unsigned integral type, if such a type exists. Since there are only two possible 8-bit unsigned integral types, char for a compiler that treats it as unsigned and unsigned char, it must be one of those. Except on systems where char is larger than 8 bits, in which case it won't exist.