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Overloading operator== with `&&` and `const` qualifier cause ambiguity in C++20

Consider the struct S with two operator== overloads of same && qualifier and different const qualifier:

struct S {
  bool operator==(const S&) && { 
    return true;
  }
  bool operator==(const S&) const && { 
    return true;
  }
};

If I compare the two S with operator==:

S{} == S{};

gcc and msvc accept this code, clang rejects it with:

<source>:14:7: error: use of overloaded operator '==' is ambiguous (with operand types 'S' and 'S')
  S{} == S{};
  ~~~ ^  ~~~

Why does clang think there is an ambiguous overload resolution here? Shouldn't the non-const one be the best candidate in this case?

Similarly, if I compare two S with the synthesized operator!=:

S{} != S{};

gcc still accept this code, but msvc and clang doesn't:

<source>:14:7: error: use of overloaded operator '!=' is ambiguous (with operand types 'S' and 'S')
  S{} != S{};
  ~~~ ^  ~~~

It seems weird that the synthesized operator!= suddenly causes the ambiguity for msvc. Which compiler is right?

like image 588
康桓瑋 Avatar asked Mar 23 '21 10:03

康桓瑋


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1 Answers

The example would be unambiguous in C++17. C++20 brings change:

[over.match.oper]

For a unary operator @ with an operand of type cv1 T1, and for a binary operator @ with a left operand of type cv1 T1 and a right operand of type cv2 T2, four sets of candidate functions, designated member candidates, non-member candidates, built-in candidates, and rewritten candidates, are constructed as follows:

  • ...
  • For the operator ,, the unary operator &, or the operator ->, the built-in candidates set is empty. For all other operators, the built-in candidates include all of the candidate operator functions defined in [over.built] that, compared to the given operator,
    • have the same operator name, and
    • accept the same number of operands, and
    • accept operand types to which the given operand or operands can be converted according to [over.best.ics], and
    • do not have the same parameter-type-list as any non-member candidate that is not a function template specialization.

The rewritten candidate set is determined as follows:

  • ...
  • For the equality operators, the rewritten candidates also include a synthesized candidate, with the order of the two parameters reversed, for each non-rewritten candidate for the expression y == x.

Thus, the rewritten candidate set includes these:

 implicit object parameter
 |||
(S&&, const S&);       // 1
(const S&&, const S&); // 2

// candidates that match with reversed arguments
(const S&, S&&);       // 1 reversed
(const S&, const S&&); // 2 reversed

The overload 1 is better match than 2, but the synthesised reversed overload of 1 is ambiguous with the original non-reversed overload because both have const conversion to one parameter. Note that this is actually ambiguous even if overload 2 doesn't exist.

Thus, Clang is correct.


This is also covered by the informative compatibility annex:

Affected subclause: [over.match.oper] Change: Equality and inequality expressions can now find reversed and rewritten candidates.

Rationale: Improve consistency of equality with three-way comparison and make it easier to write the full complement of equality operations.

Effect on original feature: Equality and inequality expressions between two objects of different types, where one is convertible to the other, could invoke a different operator. Equality and inequality expressions between two objects of the same type could become ambiguous.

struct A {
  operator int() const;
};

bool operator==(A, int);        // #1
// #2 is built-in candidate: bool operator==(int, int);
// #3 is built-in candidate: bool operator!=(int, int);

int check(A x, A y) {
  return (x == y) +             // ill-formed; previously well-formed
    (10 == x) +                 // calls #1, previously selected #2
    (10 != x);                  // calls #1, previously selected #3
}
like image 68
eerorika Avatar answered Sep 28 '22 01:09

eerorika