I want to overload the assignment operator for types like "int", "long" etc. That is, I want to use code like:
Apparently assignment operators cannot be friend functions. How do I achieve the above?
I maybe missing something simple but just cant figure out the syntax/method to do this.
Also, one IMP restriction - we CANNOT use get/set methods because :: In release code, we will have CX typedefed as int or long (as required), but in DEBUG code, we want to have it as a class (for automatic type checking in thousands of places). The code needs to be common. Reason is that the compiler (atleast the version we are using) is somehow not able to optimize all the operations if CX is made a class.
One issue is - I dont want this to pass:
I assume the casting solution below will implicitly make the code for long/short etc pass too. (If not, then it is exactly what I am looking for!).
On a related note, answering David's question - the reason I want to refactor is - we want the ability to switch CX to be 32 bit or 64 bit. As such, we want to disallow any implicit conversions and catch them at compile time. Now, the reverse - (disallowiong
CX x = some_64_bit_int;
CX x = some_32_bit_int;
I achieved by using a templatized = operator that asserts on compile time by default, but overloading it for my desired type.
In-case you feel this is a bad design or that I should try other alternatives - The reason why I need is this: We have thousands of lines of legacy code, where something is just typedefed to "int".
There are assignments all over the place like:
I am working on a project to change CX to a class. In the first phase, we want to fix all compilation errors (that come on making CX a class) making as little changes to code as possible.
A copy assignment operator of class is a non-template non-static member function with the name operator= that takes exactly one parameter of type T, T&, const T&, volatile T&, or constvolatile T&. For a type to be , it must have a public copy assignment operator.
|class_nameclass_name ( class_name )||(1)|
|class_nameclass_name ( const class_name )||(2)|
|class_nameclass_name ( const class_name ) = default;||(3)||(since C++11)|
|class_nameclass_name ( const class_name ) = delete;||(4)||(since C++11)|
- Typical declaration of a copy assignment operator when copy-and-swap idiom can be used.
- Typical declaration of a copy assignment operator when copy-and-swap idiom cannot be used (non-swappable type or degraded performance).
- Forcing a copy assignment operator to be generated by the compiler.
- Avoiding implicit copy assignment.
The copy assignment operator is called whenever selected by overload resolution, e.g. when an object appears on the left side of an assignment expression.
Implicitly-declared copy assignment operator
If no user-defined copy assignment operators are provided for a class type (struct, class, or union), the compiler will always declare one as an inline public member of the class. This implicitly-declared copy assignment operator has the form T& T::operator=(const T&) if all of the following is true:
- each direct base of has a copy assignment operator whose parameters are B or const B& or constvolatile B&;
- each non-static data member of of class type or array of class type has a copy assignment operator whose parameters are M or const M& or constvolatile M&.
Otherwise the implicitly-declared copy assignment operator is declared as T& T::operator=(T&). (Note that due to these rules, the implicitly-declared copy assignment operator cannot bind to a volatile lvalue argument.)
A class can have multiple copy assignment operators, e.g. both T& T::operator=(const T&) and T& T::operator=(T). If some user-defined copy assignment operators are present, the user may still force the generation of the implicitly declared copy assignment operator with the keyword .(since C++11)
The implicitly-declared (or defaulted on its first declaration) copy assignment operator has an exception specification as described in dynamic exception specification(until C++17)exception specification(since C++17)
Because the copy assignment operator is always declared for any class, the base class assignment operator is always hidden. If a using-declaration is used to bring in the assignment operator from the base class, and its argument type could be the same as the argument type of the implicit assignment operator of the derived class, the using-declaration is also hidden by the implicit declaration.
Deleted implicitly-declared copy assignment operator
A implicitly-declared copy assignment operator for class is defined as deleted if any of the following is true:
- has a user-declared move constructor;
- has a user-declared move assignment operator.
Otherwise, it is defined as defaulted.
A defaulted copy assignment operator for class is defined as deleted if any of the following is true:
- has a non-static data member of non-class type (or array thereof) that is const;
- has a non-static data member of a reference type;
- has a non-static data member or a direct or virtual base class that cannot be copy-assigned (overload resolution for the copy assignment fails, or selects a deleted or inaccessible function);
- is a union-like class, and has a variant member whose corresponding assignment operator is non-trivial.
Trivial copy assignment operator
The copy assignment operator for class is trivial if all of the following is true:
- it is not user-provided (meaning, it is implicitly-defined or defaulted) , , and if it is defaulted, its signature is the same as implicitly-defined(until C++14);
- has no virtual member functions;
- has no virtual base classes;
- the copy assignment operator selected for every direct base of is trivial;
- the copy assignment operator selected for every non-static class type (or array of class type) member of is trivial;
A trivial copy assignment operator makes a copy of the object representation as if by std::memmove. All data types compatible with the C language (POD types) are trivially copy-assignable.
Implicitly-defined copy assignment operator
If the implicitly-declared copy assignment operator is neither deleted nor trivial, it is defined (that is, a function body is generated and compiled) by the compiler if odr-used. For union types, the implicitly-defined copy assignment copies the object representation (as by std::memmove). For non-union class types (class and struct), the operator performs member-wise copy assignment of the object's bases and non-static members, in their initialization order, using built-in assignment for the scalars and copy assignment operator for class types.
The generation of the implicitly-defined copy assignment operator is deprecated(since C++11) if has a user-declared destructor or user-declared copy constructor.
If both copy and move assignment operators are provided, overload resolution selects the move assignment if the argument is an rvalue (either a prvalue such as a nameless temporary or an xvalue such as the result of std::move), and selects the copy assignment if the argument is an lvalue (named object or a function/operator returning lvalue reference). If only the copy assignment is provided, all argument categories select it (as long as it takes its argument by value or as reference to const, since rvalues can bind to const references), which makes copy assignment the fallback for move assignment, when move is unavailable.
It is unspecified whether virtual base class subobjects that are accessible through more than one path in the inheritance lattice, are assigned more than once by the implicitly-defined copy assignment operator (same applies to move assignment).
See assignment operator overloading for additional detail on the expected behavior of a user-defined copy-assignment operator.
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The following behavior-changing defect reports were applied retroactively to previously published C++ standards.
|DR||Applied to||Behavior as published||Correct behavior|
|CWG 2171||C++14||operator=(X&)=default was non-trivial||made trivial|