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 one unsigned and one signed variable are added (or any binary operation) both are implicitly converted to unsigned, which would in this case result in a huge result.

So it is safe in the sense of that the result might be huge and wrong, but it will never crash.