On type of a literal, unsigned negative number

Question

The C++ Primer says:

We can independently specify the signedees and the size of an integral literal. If the suffix contains a U, then the literal has an unsigned type, so a decimal, octal or hexadecimal literal with a U suffix has the smallest type of unsigned int, unsigned long or unsigned long long in which the literal's value fits

When one declares

int i = -12U;

The way i understand it is that -12 is converted to the unsigned version of itself (4294967284) and then assigned to an int, making the result a very large positive number due to rollover.

This does not seem to happen. What am i missing please?

cout << i << endl; // -12

Answer 1

12 has type int and the value 12.

12U has type unsigned int and the value 12.

-12U has type unsigned int and the value std::numeric_limits<unsigned int>::max() + 1 - 12.

int i = -12U; applies an implementation-defined conversion to convert -12U to type int.

Answer 2

You are assigning the unsigned int back to a signed int, so it gets converted again.

It's like you did this:

int i = (int)(unsigned int)(-12);

Answer 3

u effectively binds more tightly than -. You are getting -(12u).