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

Answer 1

As others have pointed out, uint8_t is streamed as an unsigned char. I've sometimes made use of bit fields from integer types that are streamed as, well, integers, to avoid having to cast or overload operator<<, but only when it doesn't waste space, like in the Pos struct below:

#include <iostream>

struct WasteAbyte {
    unsigned short has_byte_range:8;
};

struct Pos {
    unsigned short x:8;
    unsigned short y:8;
};

int main() {
    WasteAbyte W = {255};

    ++W.has_byte_range;

    std::cout << W.has_byte_range << std::endl;

    std::cout << sizeof(WasteAbyte) << std::endl;
    std::cout << sizeof(Pos) << std::endl;

    return 0;
}

Output:

0
2
2

Answer 2

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 3

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.