In the following code, it appears that std::cout
does not print the variable of type uint8_t properly.
int main() {
uint8_t var = 16;
std::cout << "value: " << var << std::endl;
}
output:
value:
I don't have problem with similar uintX_t types.
I know I'm missing something here, but I can't figure what it is.
uint8_t
maps tounsigned char
on most systems. As the result, the override that interprets16
as a non-printable character gets invoked. Add a cast toint
to see the numeric value: