Signed to unsigned conversion in C - is it always safe?

Question

Suppose I have the following C code.

unsigned int u = 1234;
int i = -5678;

unsigned int result = u + i;

What implicit conversions are goin

Answer 1

When converting from signed to unsigned there are two possibilities. Numbers that were originally positive remain (or are interpreted as) the same value. Number that were originally negative will now be interpreted as larger positive numbers.