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Only extant operators in the language may be overloaded, by defining new functions with identifiers such as "+", "*", "&" etc. Subsequent revisions of the language (in 1995 and 2005) maintain the restriction to overloading of extant operators. In C++, operator overloading is more refined than in ALGOL 68. [44]
All logical operators exist in C and C++ and can be overloaded in C++, albeit the overloading of the logical AND and logical OR is discouraged, because as overloaded operators they behave as ordinary function calls, which means that both of their operands are evaluated, so they lose their well-used and expected short-circuit evaluation property ...
In the C++ programming language, the assignment operator, =, is the operator used for assignment.Like most other operators in C++, it can be overloaded.. The copy assignment operator, often just called the "assignment operator", is a special case of assignment operator where the source (right-hand side) and destination (left-hand side) are of the same class type.
The previous section notwithstanding, there are other ways in which ad hoc polymorphism can work out. Consider for example the Smalltalk language. In Smalltalk, the overloading is done at run time, as the methods ("function implementation") for each overloaded message ("overloaded function") are resolved when they are about to be executed.
The first is taken in C++: "in C++, there is no overloading across scopes." [12] As a result, to obtain an overload set with functions declared in different scopes, one needs to explicitly import the functions from the outer scope into the inner scope, with the using keyword.
In languages that support operator overloading by the programmer (such as C++) but have a limited set of operators, operator overloading is often used to define customized uses for operators. In the example IF ORDER_DATE > "12/31/2011" AND ORDER_DATE < "01/01/2013" THEN CONTINUE ELSE STOP, the operators are: > (greater than), AND and < (less than).
Here, attempting to use a non-class type in a qualified name (T::foo) results in a deduction failure for f<int> because int has no nested type named foo, but the program is well-formed because a valid function remains in the set of candidate functions.
The move assignment operator, like most C++ operators, can be overloaded. Like the copy assignment operator it is a special member function . If the move assignment operator is not explicitly defined, the compiler generates an implicit move assignment operator ( C++11 and newer) provided that copy / move constructors , copy assignment operator ...