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

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);