How to subtract two unsigned ints with wrap around or overflow

Question

There are two unsigned ints (x and y) that need to be subtracted. x is always larger than y. However, both x and y can wrap around; for example, if they were both bytes, af

Answer 1

Assuming two unsigned integers:

• If you know that one is supposed to be "larger" than the other, just subtract. It will work provided you haven't wrapped around more than once (obviously, if you have, you won't be able to tell).
• If you don't know that one is larger than the other, subtract and cast the result to a signed int of the same width. It will work provided the difference between the two is in the range of the signed int (if not, you won't be able to tell).

To clarify: the scenario described by the original poster seems to be confusing people, but is typical of monotonically increasing fixed-width counters, such as hardware tick counters, or sequence numbers in protocols. The counter goes (e.g. for 8 bits) 0xfc, 0xfd, 0xfe, 0xff, 0x00, 0x01, 0x02, 0x03 etc., and you know that of the two values x and y that you have, x comes later. If x==0x02 and y==0xfe, the calculation x-y (as an 8-bit result) will give the correct answer of 4, assuming that subtraction of two n-bit values wraps modulo 2ⁿ - which C99 guarantees for subtraction of unsigned values. (Note: the C standard does not guarantee this behaviour for subtraction of signed values.)