I want to work with unsigned 8-bit variables in C++. Either unsigned char
or uint8_t
do the trick as far as the arithmetic is concerned (which is expected, since AFAIK uint8_t
is just an alias for unsigned char
, or so the debugger presents it.
The problem is that if I print out the variables using ostream in C++ it treats it as char. If I have:
unsigned char a = 0;
unsigned char b = 0xff;
cout << "a is " << hex << a <<"; b is " << hex << b << endl;
then the output is:
a is ^@; b is 377
instead of
a is 0; b is ff
I tried using uint8_t
, but as I mentioned before, that's typedef'ed to unsigned char
, so it does the same. How can I print my variables correctly?
Edit: I do this in many places throughout my code. Is there any way I can do this without casting to int
each time I want to print?
I would suggest using the following technique:
struct HexCharStruct
{
unsigned char c;
HexCharStruct(unsigned char _c) : c(_c) { }
};
inline std::ostream& operator<<(std::ostream& o, const HexCharStruct& hs)
{
return (o << std::hex << (int)hs.c);
}
inline HexCharStruct hex(unsigned char _c)
{
return HexCharStruct(_c);
}
int main()
{
char a = 131;
std::cout << hex(a) << std::endl;
}
It's short to write, has the same efficiency as the original solution and it lets you choose to use the "original" character output. And it's type-safe (not using "evil" macros :-))