If no operand of an operator in an expression has a type that is a class
or an enumeration, the operator is assumed to be a built-in operator
and interpreted according to [expr.compound].

[*Note 1*: *end note*]

Because
.,
.*,
and
::
cannot be overloaded,
these operators are always built-in operators interpreted according to
[expr.compound].

?:
cannot be overloaded, but the rules in this subclause are used to determine
the conversions to be applied to the second and third operands when they
have class or enumeration type ([expr.cond]).

β [*Example 1*: struct String {
String (const String&);
String (const char*);
operator const char* ();
};
String operator + (const String&, const String&);
void f() {
const char* p= "one" + "two"; // error: cannot add two pointers; overloaded operator+ not considered
// because neither operand has class or enumeration type
int I = 1 + 1; // always evaluates to 2 even if class or enumeration types exist
// that would perform the operation.
}
β *end example*]

If either operand has a type that is a class or an enumeration, a
user-defined operator function can be declared that implements
this operator or a user-defined conversion can be necessary to
convert the operand to a type that is appropriate for a built-in
operator.

In this case, overload resolution is used to determine
which operator function or built-in operator is to be invoked to implement the
operator.

Therefore, the operator notation is first transformed
to the equivalent function-call notation as summarized in
Table 19
(where @ denotes one of the operators covered in the specified subclause).

However, the operands are sequenced in the order prescribed
for the built-in operator ([expr.compound]).

Table 19: Relationship between operator and function call notation [tab:over.match.oper]

Subclause | Expression | As member function | As non-member function | |

@a | (a).operator@ ( ) | operator@(a) | ||

a@b | (a).operator@ (b) | operator@(a, b) | ||

a=b | (a).operator= (b) | |||

a[b] | (a).operator[](b) | |||

a-> | (a).operator->( ) | |||

a@ | (a).operator@ (0) | operator@(a, 0) |

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:

- If T1 is a complete class type or a class currently being defined, the set of member candidates is the result of a search for operator@ in the scope of T1; otherwise, the set of member candidates is empty.
- For the operators =, [], or ->, the set of non-member candidates is empty; otherwise, it includes the result of unqualified lookup for operator@ in the rewritten function call ([basic.lookup.unqual], [basic.lookup.argdep]), ignoring all member functions.However, if no operand has a class type, only those non-member functions in the lookup set that have a first parameter of type T1 or βreference to cv T1β, when T1 is an enumeration type, or (if there is a right operand) a second parameter of type T2 or βreference to cv T2β, when T2 is an enumeration type, are candidate functions.
- 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 relational ([expr.rel]) operators, the rewritten candidates include all non-rewritten candidates for the expression x <=> y.
- For the relational ([expr.rel]) and three-way comparison ([expr.spaceship]) 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.
- 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 that is a rewrite target with first operand y.
- For all other operators, the rewritten candidate set is empty.

A non-template function or function template F named operator==
is a rewrite target with first operand o
unless a search for the name operator!= in the scope S
from the instantiation context of the operator expression
finds a function or function template
that would correspond ([basic.scope.scope]) to F
if its name were operator==,
where S is the scope of the class type of o
if F is a class member, and
the namespace scope of which F is a member otherwise.

A function template specialization named operator== is a rewrite target
if its function template is a rewrite target.

[*Example 2*: struct A {};
template<typename T> bool operator==(A, T); // #1
bool a1 = 0 == A(); // OK, calls reversed #1
template<typename T> bool operator!=(A, T);
bool a2 = 0 == A(); // error, #1 is not a rewrite target
struct B {
bool operator==(const B&); // #2
};
struct C : B {
C();
C(B);
bool operator!=(const B&); // #3
};
bool c1 = B() == C(); // OK, calls #2; reversed #2 is not a candidate
// because search for operator!= in C finds #3
bool c2 = C() == B(); // error: ambiguous between #2 found when searching C and
// reversed #2 found when searching B
struct D {};
template<typename T> bool operator==(D, T); // #4
inline namespace N {
template<typename T> bool operator!=(D, T); // #5
}
bool d1 = 0 == D(); // OK, calls reversed #4; #5 does not forbid #4 as a rewrite target
β *end example*]

For the built-in assignment operators, conversions of the left
operand are restricted as follows:

- no temporaries are introduced to hold the left operand, and
- no user-defined conversions are applied to the left operand to achieve a type match with the left-most parameter of a built-in candidate.

The set of candidate functions for overload resolution
for some operator @
is the
union of
the member candidates,
the non-member candidates,
the built-in candidates,
and the rewritten candidates
for that operator @.

The argument list contains all of the
operands of the operator.

The best function from the set of candidate functions is selected
according to [over.match.viable]
and [over.match.best].112

[*Example 3*: struct A {
operator int();
};
A operator+(const A&, const A&);
void m() {
A a, b;
a + b; // operator+(a, b) chosen over int(a) + int(b)
}
β *end example*]

If a rewritten operator<=> candidate
is selected by overload resolution
for an operator @,
x @ y
is interpreted as
0 @ (y <=> x)
if the selected candidate is a synthesized candidate
with reversed order of parameters,
or (x <=> y) @ 0 otherwise,
using the selected rewritten operator<=> candidate.

Rewritten candidates for the operator @
are not considered in the context of the resulting expression.

If a rewritten operator== candidate
is selected by overload resolution
for an operator @,
its return type shall be cv bool, and
x @ y is interpreted as:

- if @ is != and the selected candidate is a synthesized candidate with reversed order of parameters, !(y == x),
- otherwise, if @ is !=, !(x == y),
- otherwise (when @ is ==), y == x,

If a built-in candidate is selected by overload resolution, the
operands of class type are converted to the types of the corresponding parameters
of the selected operation function, except that the second standard conversion
sequence of a user-defined conversion sequence is not applied.

Then the operator is treated as the corresponding
built-in operator and interpreted according to [expr.compound].

[*Example 4*: struct X {
operator double();
};
struct Y {
operator int*();
};
int *a = Y() + 100.0; // error: pointer arithmetic requires integral operand
int *b = Y() + X(); // error: pointer arithmetic requires integral operand
β *end example*]

If the operator is the operator
,,
the unary operator
&,
or the operator
->,
and there are no viable functions, then the operator is
assumed to be the built-in operator and interpreted according to
[expr.compound].

[*Note 3*: *end note*]

The lookup rules for operators in expressions are different than
the lookup
rules for operator function names in a function call, as shown in the following
example:
struct A { };
void operator + (A, A);
struct B {
void operator + (B);
void f ();
};
A a;
void B::f() {
operator+ (a,a); // error: global operator hidden by member
a + a; // OK, calls global operator+
}

β