[ Note

: *end note*

]Operators can be overloaded, that is, given meaning when applied to
expressions of class type or
enumeration type.

Uses of overloaded operators are transformed into
function calls as described in [over.oper].

Overloaded operators
obey the rules for syntax and evaluation order specified in [expr.compound],
but the requirements of operand type and value category are replaced
by the rules for function call.

— Subclause [expr.compound] defines the effects of operators when applied to types
for which they have not been overloaded.

Operator overloading shall not
modify the rules for the *built-in operators*,
that is, for operators applied to types for which they are defined by this
Standard.

However, these built-in operators participate in overload
resolution, and as part of that process user-defined conversions will be
considered where necessary to convert the operands to types appropriate
for the built-in operator.

If a built-in operator is selected, such
conversions will be applied to the operands before the operation is
considered further according to the rules in subclause [expr.compound];
see [over.match.oper], [over.built].

If during the evaluation of an expression, the result is not
mathematically defined or not in the range of representable values for
its type, the behavior is undefined.

[ Note

: *end note*

]The implementation may regroup operators according to
the usual mathematical rules only
where the operators really are associative or commutative.50

For example, in the following fragment

int a, b; /* ... */ a = a + 32760 + b + 5;the expression statement behaves exactly the same as

a = (((a + 32760) + b) + 5);due to the associativity and precedence of these operators.

Thus, the
result of the sum (a + 32760) is next added to b, and
that result is then added to 5 which results in the value assigned to
a.

On a machine in which overflows produce an exception and in
which the range of values representable by an int is
[-32768, +32767], the implementation cannot rewrite this
expression as

a = ((a + b) + 32765);since if the values for a and b were, respectively, -32754 and -15, the sum a + b would produce an exception while the original expression would not; nor can the expression be rewritten as either

a = ((a + 32765) + b);or

a = (a + (b + 32765));since the values for a and b might have been, respectively, 4 and -8 or -17 and 12.

However on a machine in which
overflows do not produce an exception and in which the results of
overflows are reversible, the above expression statement can be
rewritten by the implementation in any of the above ways because the
same result will occur.

— The cast and assignment operators must still perform their specific
conversions as described in [expr.type.conv], [expr.cast],
[expr.static.cast] and [expr.ass].

⮥