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?
In Python, overloading is achieved by overriding the method which is specifically for that operator, in the user-defined class. For example, __add__(self, x) is a method reserved for overloading + operator, and __eq__(self, x) is for overloading == .
Overloading operator == does not give you operator !=
Explanation: ?:, :: and . cannot be overloaded whereas == can be overloaded.
[13.6] Can I overload operator== so it lets me compare two char[] using a string comparison? No: at least one operand of any overloaded operator must be of some class type.
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 }
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