On 2011-05-28 21:59:57 +0300, Matt Calabrese said:
On Sat, May 28, 2011 at 2:55 PM, Steven Watanabe <watanabesj@gmail.com>wrote:
I don't have the concepts proposal handy, but doesn't LessThanComparable have an axiom that requires that (a < b) is equivalent to (b > a)?
Yes.
axiom Consistency(T a, T b) { (a > b) == (b < a); (a <= b) == !(b < a); (a >= b) == !(a < b); }
For what it's worth, e.g. NaN's are an exception to that rule... #include <limits> #include <iostream> #define PRINT(x) do { std::cout << #x ": " << (x) << std::endl; } while(0) int main() { double one = 1; double nan = std::numeric_limits<double>::quiet_NaN(); PRINT(one == nan); // false PRINT(one != nan); // true PRINT(one < nan); // false PRINT(one > nan); // false PRINT(one <= nan); // false PRINT(one >= nan); // false } -- Pyry Jahkola (a.k.a. pyrtsa)