Proposed resolution (approved by CWG 2024-03-20):

The operator function with parameters x and y is defined as deleted if

• a first overload resolution (12.2 [over.match]), as applied to x @ y,
  • does not result in a usable candidate, or
  • the selected candidate selected by overload resolution is not a rewritten candidate. , or
• a second overload resolution for the expression resulting from the interpretation of x @ y using the selected rewritten candidate (12.2.2.3 [over.match.oper]) does not result in a usable candidate (for example, that expression might be (x <=> y) @ 0), or
• x @ y cannot be implicitly converted to bool.

In any of the two overload resolutions above, the defaulted operator function is not considered as a candidate for the @ operator. Otherwise, the operator function yields x @ y. The defaulted operator function is not considered as a candidate in the overload resolution for the @ operator.

[Accepted as a DR at the March, 2024 meeting.]

(See also editorial issues 5335 and 5336.)

Consider the example in 11.10.4 [class.compare.secondary] paragraph 3:

  struct HasNoLessThan { };
  struct C {
    friend HasNoLessThan operator<=>(const C&, const C&);
    bool operator<(const C&) const = default;  // OK, function is deleted
  };

While the comment may reflect the intent, it does not follow from the wording. 11.10.4 [class.compare.secondary] paragraph 2 specifies:

The operator function with parameters x and y is defined as deleted if

• overload resolution (12.2 [over.match]), as applied to x @ y, does not result in a usable candidate, or
• the candidate selected by overload resolution is not a rewritten candidate.

Otherwise, the operator function yields x @ y. The defaulted operator function is not considered as a candidate in the overload resolution for the @ operator.

Overload resolution applied to x < y results in a usable candidate operator<=> (12.2.1 [over.match.general]) and that candidate is a rewritten candidate (12.2.2.3 [over.match.oper] bullet 3.4), thus operator< in the above example is not deleted. However, its definition is ill-formed, because the rewrite (x <=> y) < 0 is ill-formed (12.2.2.3 [over.match.oper] paragraph 8).

There is implementation divergence.

Subclause 11.10.3 [class.spaceship] paragraph 1 seems to prefer an ill-formed program for similar synthesized situations:

[Note 1: A synthesized three-way comparison is ill-formed if overload resolution finds usable candidates that do not otherwise meet the requirements implied by the defined expression. —end note]