A few times I ran into the following issue
uint8_t a = 0;
std::cout << "a = " << a << std::endl;
Guess what would it print?
a =
What happened here is that it printed NULL (0) in ASCII. uint8_t is just a typedef of unsigned char.
typedef unsigned char uint8_t; //https://code.woboq.org/gtk/include/stdint.h.html#uint8_t
In the past, I worked around the problem by casting the variable to e.g. uint16_t and be done with it. Then I found out this:
uint8_t a = 0;
std::cout << "a = " << +a << std::endl;
a = 0
You can print +a, which would promote the type to an integral type. Then you get 0 printed out.
According to the reference (https://en.cppreference.com/w/cpp/language/operator_arithmetic),
The built-in unary plus operator returns the value of its operand. The only situation where it is not a no-op is when the operand has integral type or unscoped enumeration type, which is changed by integral promotion, e.g, it converts char to int or if the operand is subject to lvalue-to-rvalue, array-to-pointer, or function-to-pointer conversion.