Proposed resolution:

This wording is relative to N4950.

1. Modify [container.opt.reqmts] as indicated:

   a <=> b
   

   -2- Result: synth-three-way-result<X::value_type>.

   -3- Preconditions: Either <=> is defined for values of type (possibly const) T models three_way_comparable, or < is defined for values of type (possibly const) T and < is a total ordering relationship.

   -4- Returns: lexicographical_compare_three_way(a.begin(), a.end(), b.begin(), b.end(), synth-three-way)

   [Note 1: The algorithm lexicographical_compare_three_way is defined in Clause 27. — end note]

   -5- Complexity: Linear.

[ 2023-11-11 Approved at November 2023 meeting in Kona. Status changed: Voting → WP. ]

[ 2023-06-14 Varna; Move to Ready ]

[ 2023-06-13; Varna ]

The group liked the previously suggested wording but would prefer to say "models" instead of "satisfies" in preconditions.

[ 2022-04-24; Daniel rebases wording on N4910 ]

[ 2020-04-25 Issue Prioritization ]

Priority to 2 after reflector discussion.

Previous resolution [SUPERSEDED]:

This wording is relative to N4861.

1. Modify [container.requirements.general], Table [tab:container.opt], as indicated:

   Table 75: Optional container operations [tab:container.opt]

   Expression	Return type	Operational semantics	Assertion/note pre-/post-condition	Complexity
   …
   a <=> b	synth-three-way-result<value_type>	lexicographical_compare_three_way( a.begin(), a.end(), b.begin(), b.end(), synth-three-way)	Preconditions: Either <=> is defined for values of type (possibly const) T satisfies three_way_comparable, or < is defined for values of type (possibly const) T and < is a total ordering relationship.	linear
   …

The precondition for <=> on containers is:

"Either <=> is defined for values of type (possibly const) T, or < is defined for values of type (possibly const) T and < is a total ordering relationship."

I don't think <=> is sufficient, because synth-three-way won't use <=> unless three_way_comparable<T> is satisfied, which requires weakly-equality-comparable-with<T, T> as well as <=>.

So to use <=> I think the type also requires ==, or more precisely, it must satisfy three_way_comparable.

The problem becomes clearer with the following example:

#include <compare>
#include <vector>

struct X
{
  friend std::strong_ordering operator<=>(X, X) { return std::strong_ordering::equal; }
};

std::vector<X> v(1);
std::strong_ordering c = v <=> v;

This doesn't compile, because despite X meeting the preconditions for <=> in [tab:container.opt], synth-three-way will return std::weak_ordering.

Here is another example:

#include <compare>
#include <vector>

struct X
{
  friend bool operator<(X, X) { return true; } // The return value is intentional, see below
  friend std::strong_ordering operator<=>(X, X) { return std::strong_ordering::equal; }
};

std::vector<X> v(1);
std::weak_ordering c = v <=> v;

This meets the precondition because it defines <=>, but the result of <=> on vector<X> will be nonsense, because synth-three-way will use operator< not operator<=> and that defines a broken ordering.

So we're stating a precondition which implies "if you do this, you don't get garbage results" and then we give garbage results anyway.

The proposed resolution is one way to fix that, by tightening the precondition so that it matches what synth-three-way actually does.