6 ms·
Maybe you can explain something that puzzles me - why is (1 / 0) * (1 / (1 / 0)) equal to NaN? Edit: just checked and haskell behaves the same way.
by dev360 12y ago
Maybe you can explain something that puzzles me - why is (1 / 0) * (1 / (1 / 0)) equal to NaN?
Edit: just checked and haskell behaves the same way.
- CHY872 12y agoInfinity * 0 returns NaN.