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 <cstdint>
#include <iostream>
#include <limits>

int main()
{
    uint8_t x = 12;
    std::cout << (x << 1) << '\n';
    std::cout << ~x;

    std::cin.clear();
    std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
    std::cin.get();

    return 0;
}

My output is the following.

24
-13

I tested larger numbers and operator << always gives me positive numbers, while operator ~ always gives me negative numbers. I then used sizeof() and found...

When I use the left shift bitwise operator(<<), I receive an unsigned 4 byte integer.

When I use the bitwise not operator(~), I receive a signed 4 byte integer.

It seems that the bitwise not operator(~) does a signed integral promotion like the arithmetic operators do. However, the left shift operator(<<) seems to promote to an unsigned integral.

I feel obligated to know when the compiler is changing something behind my back. If I'm correct in my analysis, do all the bitwise operators promote to a 4 byte integer? And why are some signed and some unsigned? I'm so confused!

Edit: My assumption of always getting positive or always getting negative values was wrong. But from being wrong, I understand what was really happening thanks to the great answers below.

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).