4 ms·

Your compiler will take care of that. Leave the division for the humans to read.

• xaduha 5y ago
  I'm in a camp that thinks compilers should also take care of the original unsigned average(unsigned a, unsigned b) { return (a + b) / 2; } At the end of the day it's all just text. There are plenty of steps before any of it does anything at all.
  • Dylan16807 5y ago
    What should happen if you store "a+b" in an intermediate value?
    • xaduha 5y ago
      If it is used and there's no way around it, then show a compilation warning that there might be overflow. If it can be resolved without being directly used, then it should be optimized away.
  • jws 5y ago
    For C at least, the spec says that unsigned addition is modulo 2^64 (or 32 or 16 or whatever) so, imagine you had an 8 bit unsigned, 128+128 gives you 0. Divided by 2 is 0. That’s the right answer by the language specification. The trick is to get 128.