idea1434

3 ms·

1. 1.▲
   by idea1434 9y ago

   It just makes more sense. Similar to how a<=b is equivalent to (a<b)&&(a==b). Coming up with an entirely fresh meaning for <=> involving spaceships and a conglomerated comparative threesome just seems illogical to me, gi
2. 2.▲
   by idea1434 9y ago

   No, it would be the same as evaluating (a<b)&&(a==b)&&(a>b), not std::is_nan.
3. 3.▲
   by idea1434 9y ago

   No, it could be used for any type. It would be equivalent to (a<b)&&(a==b)&&(a>b), but would be most useful for floating point, where this operator would evaluate to false if a or b is NaN.
4. 4.▲
   by idea1434 9y ago

   I would like to see a spaceship operator for floating point comparison, where a<=>b means a is less than, equal to, or greater than b. It would be useful for checking that a and b aren't NaN. If either were NaN, then the comparis