What is going on with bitwise operators and integer promotion?

Question

I have a simple program. Notice that I use an unsigned fixed-width integer 1 byte in size.

#include 
#include 

        

Answer 1

[expr.unary.op]

The operand of ~ shall have integral or unscoped enumeration type; the result is the one’s complement of　its operand. Integral promotions are performed.

[expr.shift]

The shift operators << and >> group left-to-right. [...] The operands shall be of integral or unscoped enumeration type and integral promotions are performed.

What's the integral promotion of uint8_t (which is usually going to be unsigned_char behind the scenes)?

[conv.prom]

A prvalue of an integer type other than bool, char16_t, char32_t, or wchar_t whose integer conversion rank (4.13) is less than the rank of int can be converted to a prvalue of type int if int can represent all the values of the source type; otherwise, the source prvalue can be converted to a prvalue of type unsigned int.

So int, because all of the values of a uint8_t can be represented by int.

What is int(12) << 1 ? int(24).

What is ~int(12) ? int(-13).