Question on predefined meanings for operators - c++

In "The C++ programming language", at page 265, the author makes the following statement:
Because of historical accident, the operators = (assignment), & (address-of), and , (sequencing;
§6.2.2) have predefined meanings when applied to class objects. These predefined meanings can
be made inaccessible to general users by making them private:
Then the following example is given:
class X {
private:
void operator=(const X&);
void operator&();
void operator,(const X&);
// ...
};
void f(X a, X b)
{
a = b; // error: operator= private
&a; // error: operator& private
a,b; // error: operator, private
}
I can't quite understand what do these "error" comments refer to? Does that mean I should not define a function like f, or that all of the =, &, and , operators should be used according to the default way, and it is not necessary to redefine them?

This example simply shows a way to prevent yourself or other developers of the code from using operators, which can be used without having been defined in the class, because they're automatically generated (and have default meanings for the operations they represent).
The author of the example meant, that if you try to assign b to a (in line a = b) it will cause an error, because the assignment operator is private in the class definition.
Similar error occurs in case of address-of in the second line, and the comma operator in the third.
Making default operators/constructors private if you know they're not supposed to be used (or haven't been implemented yet) is good, because one may accidentally use a very frequent operator like assignment or copy-constructor, being unaware that it's default behavior conflicts with the class lifecycle. If such operator or constructor is made private at the very beginning of class design, the compiler will generate a compile-time error instead of performing a potentially dangerous operation without notice if the programmer accidentally uses the method.
Think default assignment operator and member pointer: it will copy the pointer whereas you might want the object to be the owner of data. Then, after someone assigns one object to another without knowing that assignment is not implemented, you will end up with a double free error. Instead of that, if the operator is private, you'll get a nice error and the code will not even compile, and you'll know what's going on.

The author intends to point out here that the operators =, & and , are usually implicitly available for a class.
So if you don't want your objects to be operated on through them then you declare them as private thus disallowing their use.
Since they are declared as private you cannot access them anymore outside the class and the compiler gives you a compilation error. The function is an example showing that.

Providing your own implementation of any operator is basically the same as implementing a class method. Operators and methods are the same in terms of accessibility. What you do is disallowing access to operators from the caller's code.
It's absolutely the same as if you defined a private method and then tried to call it from some code that is not part of your class. Just make the operators public and errors will go away.

It basically prevents any one from making an 'X' object and using the "=", "&", and "," operators on that class. Because the author of the class may implement those objects with a meaning that is quite different to what the consumer of the class might think they do ... so its best to prevent them being used at all in the case of ambiguity.

The function f is an example of a user trying to use the private operators. It shows you what code it is that you're preventing by making them private. The comment // error means that a program that contained that line would fail to compile for the stated reason.

Before discussing the error, a key here is to understand that these operations will be implicitly made available for your class. This is the essence of Scott Meyers' advice "Know what functions C++ silently writes and calls."
C++ will automatically implement the assignment operator for your class, but it may not be done correctly (for example, if your class contains a pointer member variable). By defining the assignment operator explicitly, you are telling the compiler to use your implementation instead of generating one for you. And by making it private, you are essentially disallowing assignment of one class instance to another. Anywhere you try to do this in your code, the compiler will complain, which is a good thing if you really don't want assignment to be done.
In function f the author is showing you that these statements will not compile because of how the operators are defined in the class. It is perfectly acceptable to redefine operators for your class, and sometimes it is definitely required (for example, to implement a deep copy of a pointer member variable in your class). The point of the example is to show that a) you can provide your own implementation of these operators for your class, and b) because of this you have control over whether the operators are supported and implemented correctly for your class.

Related

Can copy-constructor with delete qualifier be private? [duplicate]

In order to make an object non-copiable we can explicitly delete both its copy-constructor and copy-assignment operator.
My question is: What is the right place to do it - in the public, private or protected section of the class? And - does this choice make any difference?
what is the right place to do it - in the public, private or protected section of the class?
I would put them in the public section.
This is because deleting a constructor or an assignment operator is orthogonal to making them private / protected; and when these aren't deleted, they are public by default. Putting the deletions in one of those two sections seems to me like hinting "If I hadn't deleted them, I would have made them private/protected" - which is not a message you want to convey in your case.
Note, though, that the compiler doesn't care which section you put the deletion in.
Does where we put the deleted definition make any difference?
From a pure language standpoint it makes absolutely zero difference. Name lookup and overload resolution happen before access checking. And attempting to refer to a deleted function at the end of overload resolution makes your program ill-formed, period. A compiler may or may not issue another diagnostic about the accessibility, but the program already has an error that must be reported.
So you can put that deleted definition with whatever accessibility you desire. I think most will keep it private, to be inline with the "old" practice of making a class non-copyable (put the declaration of those members in the private section of the class, and not define them), if only to help those who know the old ways "get it" sooner. A mixture of idioms, if you would.
Marking as private is also something you can't avoid if you need to support both C++03 and C++11 mode. With the help of a macro, a header can be made to conform to both standards easily:
#if __cplusplus >= 201103L
#define DELETED_DEFINITION = delete
#else
#define DELETED_DEFINITION
#endif
class noncopyable {
private:
// This header can be compiled as both C++11 and C++03
noncopyable(noncopyable const&) DELETED_DEFINITION;
void operator=(noncopyable const&) DELETED_DEFINITION;
};
From Scott Meyers's book, Effective Modern C++ (Item 10), it seems that it is better to define them as public:
By convention, deleted functions are declared public, not private.
There’s a reason for that. When client code tries to use a member
function, C++ checks accessibility before deleted status. When client
code tries to use a deleted private function, some compilers complain
only about the function being private, even though the function’s
accessibility doesn’t really affect whether it can be used. It’s worth
bearing this in mind when revising legacy code to replace
private-and-not-defined member functions with deleted ones, because
making the new functions public will generally result in better error
messages.
In addition, I believe that a deleted copy constructor/assignment, should be part of the class interface to be shared with ALL of the class users. Such kind of information should not be kept as secret by making them private.
delete works just as well with private access.
The effect of delete is to cause an error if the function is chosen by overload resolution.
The effect of private is to cause an error if the function is chosen by overload resolution from outside the class or its friends.
If both errors apply, the ultimate outcome is the same either way, but public might help avoid compiler messages about access privileges, which could cause confusion.
The access of a deleted function is irrelevant. In fact, for class members, it would have made more sense to add an additional access specifier (delete:). I suspect the reason they didn't do that, was that it wouldn't work for non-member functions.
For things like the copy constructor, it makes more sense stylistically to put it in the public section. The fact that a class doesn't have a copy constructor is a pretty major fact to know about the interface to the class.
For internal functions where you are declaring a particular overload as deleted in order to get compiler-time detection of an error, it makes sense to declare the function in the same section as all the other overloads.

Operator overloading in Fortran with no object creation such as in friend functions (C++)

I would like to know if there is anything similar to a "friend function", from C++, in Fortran. I would like to be able to make operator overloading without creating new objects since it is too expensive. I have already tried creating objects in a module and trying to make the overloaded operator return it, but it was not successful.
Your understanding of what friend means in C++ is incorrect. That deals with the accessibility of private class members inside the friend function - not with object creation as part of operator overloading. Fortran's accessibility model is different - any procedure defined in the same module as a type is a "friend" and there is nothing equivalent, in accessibility terms, to a member function.
The semantics of expression evaluation in both languages requires that the source implementation of an operator create a new object to store the result of the operation. Fancy compiler optimisation may prevent the actual creation of a temporary for a function result in some cases, but that very much depends on details.

Friend functions and their relationship to the I/O Operators

I've pored over information regarding friend functions and their use. They're able to access encapsulated data within a class while not breaking one of the golden rules of OOP. In purveying various source code for overloading the I/O operators (A basic operation, one of the first taught in learning C++) every one defines the operator as a friend as implements it outside of the class. My question is: does this need to be done? Why not just declare the function as a public member of the class and insert/display data from the class while keeping everything encapsulated? It seems no different than overloading other operators, yet it is a supposedly traditional approach to overloading the I/O operators.
Thanks for your time.
Let's say you want to overload operator<< for your class X, so you can use it like this:
X x;
std::cout << x;
Notice that std::cout is the first operand of the operator. To implement this as a member function, it would have to be a member of std::basic_ostream, which is the type of std::cout. You can't add members to an already defined class. That's why we declare it as a free function instead.
If you overloaded operator<< as a member of X, it would be taking an X object as its first operand, so you would use it like this:
X x;
x << something;
This is obviously not what you want when dealing with I/O.
If you have an overloaded operator like: a # b implemented as a member function, that call is translated to a.operator#(b);. That means the function must be a member of the class that's the type of the left operand. In the case of iostreams, that would be all the operators needed to be members of the iostream itself.
While iostreams do provide some insertion/extraction operators as members, you normally want to be able to add more without modifying the iostream class itself1. To do that, you pretty much need to implement the operator as a free function instead of a member function. Since you normally still want it to have access to the private parts of whatever type you're planning to read/write (insert/extract, if you prefer) it typically has to be a friend of that class.
This is an example of the so-called open/closed principle: the class should be open to extension, but closed to modification. In other words, you want to extend it without modifying it.

Opt-out of copy constructor

This might be a silly question, but...
I've been writing a number of classes that utilize non-copyable members. These classes are never initialized via the copy constructor in my source. When I try to compile without supplying my own copy-constructor, g++ throws out many errors about how it can't build a default copy constructor, due to the non-copyable member objects.
Is there a way to tell the compiler to just not give me a copy constructor?
EDIT: Yeah... feels silly... I had a case where I was invoking the copy-constructor by accident in a boost::bind call. Lesson learned.
The usual way to make things noncopyable is to declare but not define a copy constructor, and make it private so nothing can call it.
The next revision of the language will provide an explicit way to suppress these generated functions.
If you don't actually cause the copy-constructor to be called then it is not an error if the compiler would be unable to generate one. It sounds like you are (possibly indirectly) causing the copy-constructor to be used.
You can suppress the compiler generated one by declaring your own copy-constructor (you don't need to define it if you're not using it). You can place it in the private section of your class.
If this changes the error to say that the copy-constructor is inaccessible or you get link errors then you really are causing the copy-construtor to be used and you need to analyze why this is.
Not in the current version of C++. In C++ 0x, there will be an =delete; syntax to tell it that you don't want one of the special member functions the compiler will generate by default if you don't defined one yourself.
Until the new C++ 0x standard is fully supported, the best you can do is to delclare a version of the special member function, but not implement them. Normally they are made private (to help make it clear that they shouldn't be used).
Class foo
{
// ... rest of definition
private:
foo (const foo& rhs); // Do Not Implement
const foo& operator= (const foo& rhs); // Do Not Implement
};
No :)
If you want your class to be non-copyable use something like boost::noncopyable
class MyClass : private boost::noncopyable
{
}
or use a parametrizied macro in your class definition that declares a private copy constructor.

Macros to disallow class copy and assignment. Google -vs- Qt

To disallow copying or assigning a class it's common practice to make the copy constructor
and assignment operator private. Both Google and Qt have macros to make this easy and visible.
These macros are:
Google:
#define DISALLOW_COPY_AND_ASSIGN(TypeName) \
TypeName(const TypeName&); \
void operator=(const TypeName&)
Qt:
#define Q_DISABLE_COPY(Class) \
Class(const Class &); \
Class &operator=(const Class &);
Questions:
Why are the signatures of the two assignment operators different? It seems like the Qt version is correct.
What is the practical difference between the two?
It doesn't matter. The return type is not part of a function's signature, as it does not participate in overload resolution. So when you attempt to perform an assignment, both declarations will match, regardless of whether you use the return type.
And since the entire point in these macros is that the functions will never get called, it doesn't matter that one returns void.
I'd just like to mention that there is an alternative strategy for implementing an abstraction for disallowing copy and assignment of a class. The idea is to use inheritance instead of the preprocessor. I personally prefer this approach as I follow the rule of thumb that it is best to avoid using the preprocessor when at all possible.
boost::noncopyable is an example implementation. It is used as follows:
class A : noncopyable
{
...
};
See Boost.Utility, specifically boost::noncopyable. It's not a macro but a base class with private copy and assignment. It prevents the compiler from generating implicit copy and assignment in derived classes.
edit: Sorry, this was not an answer to the original question. By the way, boost::noncopyable uses a const reference as return type for the assignment operator. I was under the impression that the type of the return value doesn't matter since it's not supposed to be used. Still, making the operator private doesn't prevent usage inside the class or friends in which case a non-usual return type (like void, a const reference, etc) might lead to compilation errors and catch additional bugs.
There's no practical difference. The assignment operator signatures differ just as a matter of style. It's usual to have an assignment operator returning a reference to allow chaining:
a = b = c;
but a version returning void is also legal and will work just fine for cases when the only purpose is to just declare the operator private and therefore prohibited to use.
From the standard, 12.8, clause 9: "A user-declared copy assignment operator X::operator= is a non-static non-template member function of class X with exactly one parameter of type X, X&, const X&, volatile X&, or const volatile X&." It says nothing about the return type, so any return type is permissible.
Clause 10 says "If the class definition does not explicitly declare a copy assignment operator, one is declared implicitly."
Therefore, declaring any X::operator=(const X&) (or any other of the specified assignment types) is sufficient. Neither the body nor the return type is significant if the operator will never be used.
Therefore, it's a stylistic difference, with one macro doing what we'd likely expect and one saving a few characters and doing the job in a way that's likely to surprise some people. I think the Qt macro is better stylistically. Since we're talking macro, we're not talking about the programmer having to type anything extra, and failing to surprise people is a good thing in a language construct.
Others have already answered why it's legal to have different return values for operator=; IMHO jalf said it best.
However, you might wonder why Google uses a different return type, and I suspect it's this:
You don't have to repeat the type name when disabling the assignment operator like this. Usually the type name is the longest part of the declaration.
Of course, this reason is void given that a macro is used but still - old habits die hard. :-)
Both serve the same purpose
Once you write this one:
Class &operator=(const Class &);
you will get the benefits of chain assignments. But in this case you want the assignment operator to be private. so it doesn't matter.
Qt version is backward compatible, while google's is not.
If you develop your library and deprecate the use of assignment before you completely remove it, in Qt it will most likely retain the signature it originally had. In this case older application will continue to run with new version of library (however, they won't compile with the newer version).
Google's macro doesn't have such a property.
As several other answers have mentioned, the return type of the function doesn't participate in the function signature, so both declarations are equivalent as far as making the assignment operator unusable by clients of the class.
Personally I prefer the idiom of having a class privately inherit from an empty non-copyable base class (like boost::noncopyable, but I have my own so I can use it in projects that don't have boost available). The empty base class optimization takes care of making sure there's zero overhead, and it's simple, readable, and doesn't rely on the dreaded preprocessor macro functionality.
It also has the advantage that copy and assignment can't even be used within class implementation code - it'll fail at compile time while these macros will fail at link time (likely with a less informative error message).
Incidentally, if you have access to the Boost libraries (You don't? Why the heck not??), The Utility library has had the noncopyable class for a long time:
class YourNonCopyableClass : boost::noncopyable {
Clearer IMHO.
In practice I would say that both should not be used anymore if you have a C++11 compiler.
You should instead use the delete feature , see here
Meaning of = delete after function declaration
and here
http://www.stroustrup.com/C++11FAQ.html#default
Why : essentially because compiler message is much more clearer. When the compiler need one of the copy or copy assignment operator, it immediately points out to the line where the =delete was coded.
Better and complete explanations can also be found in Item 11: Prefer deleted functions to private undefined ones from Effective Modern C++ book by Scott Meyers