Behavior of cout << hex with uint8 and uint16

Question

I'm noticing that cout << hex is giving me strange results, and I cannot find anywhere that answers why. What I am doing is simply assigning some values to both a uint8_t and uint16_t and then attempting to write them to stdout. When I run this:

uint8_t a = 0xab;
uint16_t b = 0x24de;
cout << hex << a << endl;
cout << hex << b << endl;

That I get the result:

$./a.out

24de
$

with no value displayed for the uint8_t. What could be causing this? I didn't think there wouldn't be a cout implementation for one type for not the other.

Answer 1

std::uint8_t is an alias for unsigned char:

typedef unsigned char uint8_t;

So the overload of the inserter that takes a char& is chosen, and the ASCII representation of 0xab is written, which could technically vary by your operating system, as 0xab is in the range of Extended ASCII.

You have to cast it to an integer:

std::cout << std::hex << static_cast<int>(a) << std::endl;

Answer 2

The other answers are correct about the reason. The simplest fix is:

cout << hex << +a << endl;

Demonstration: http://ideone.com/ZHHAHX

It works because operands undergo integer promotion.