Comparing vectors for inequality use only equality operator on the vector elements. Why?

Question

Both my compilers (g++ and clang) won't compile this:

#include <vector>

struct A {
  friend bool operator!=(A const& a1, A const& a2) { return false; }
};

int main()
{
  std::vector<A> v1, v2;
  return (v1 != v2);
}

The error being that !(*__first1 == *__first2) somewhere in stl_algobase.h is invalid.

In other words, it is completely ignoring the existing operator!= of A. Needless to say that if I define an operator== then it compiles and works.

Is this how it should be according to the standard?

If so, why?

Answer 1

It's because the comparison operators want an EqualityComparable or LessThanComparable type.

With only == and <, you can derive equivalent !=, <=, >=, and >. In other words, by only implementing 2 operators, you can get all 6 comparisons (assuming I haven't made a mistake in the logic):

(a != b) is !(a == b)
(a <= b) is !(b < a)
(a >= b) is !(a < b)
(a >  b) is  (b < a)

Standard containers use this usually, and will use operator== and operator< when doing comparisons for types.

So yes, this is as it should be.

As for the second part of the question (the why), I'm actually not entirely sure why the other operators aren't used if available.