Why it is not allowed to overload "=" using friend function?
I have written a small program but it is giving error.
class comp
{
int real;
int imaginary;
public:
comp(){real=0; imaginary=0;}
void show(){cout << "Real="<<real<<" Imaginary="<<imaginary<<endl;}
void set(int i,int j){real=i;imaginary=j;}
friend comp operator=(comp &op1,const comp &op2);
};
comp operator=(comp &op1,const comp &op2)
{
op1.imaginary=op2.imaginary;
op1.real=op2.real;
return op1;
}
int main()
{
comp a,b;
a.set(10,20);
b=a;
b.show();
return 0;
}
The compilation gives the following error :-
[root#dogmatix stackoverflow]# g++ prog4.cpp
prog4.cpp:11: error: ‘comp operator=(comp&, const comp&)’ must be a nonstatic member function
prog4.cpp:14: error: ‘comp operator=(comp&, const comp&)’ must be a nonstatic member function
prog4.cpp: In function ‘int main()’:
prog4.cpp:25: error: ambiguous overload for ‘operator=’ in ‘b = a’
prog4.cpp:4: note: candidates are: comp& comp::operator=(const comp&)
prog4.cpp:14: note: comp operator=(comp&, const comp&)
Because if you do not declare it as a class member compiler will make one up for you and it will introduce ambiguity.
The assignment operator is explicitly required to be a class member operator. That is a sufficient reason for the compiler to fail to compile your code. Assignment is one of the special member functions defined in the standard (like the copy constructor) that will be generated by the compiler if you do not provide your own.
Unlike other operations that can be understood as external to the left hand side operator, the assignment is an operation that is semantically bound to the left hand side: modify this instance to be equal to the right hand side instance (by some definition of equal), so it makes sense to have it as an operation of the class and not an external operation. On the other hand, other operators as addition are not bound to a particular instance: is a+b an operation of a or b or none of them? -- a and b are used in the operation, but the operation acts on the result value that is returned.
That approach is actually recommended and used: define operator+= (that applies to the instance) as a member function, and then implement operator+ as a free function that operates on the result:
struct example {
example& operator+=( const example& rhs );
};
example operator+( const example& lhs, const example& rhs ) {
example ret( lhs );
ret += rhs;
return ret;
}
// usually implemented as:
// example operator+( example lhs, const example& rhs ) {
// return lhs += rhs; // note that lhs is actually a copy, not the real lhs
//}
Assignment(=) operator is a special operator that will be provided by the constructor to the class when programmer has not provided(overloaded) as member of the class.(like copy constructor).
When programmer is overloading = operator using friend function, two = operations will exists:
1) compiler is providing = operator
2) programmer is providing(overloading) = operator by friend function.
Then simply ambiguity will be created and compiler will gives error. Its compilation error.
There is no good reason for that, I think Stepanov proposed that there should be a free operator= and many good stuff can be done with that (even more than what can be done with the move assignment). I can't find the citation but Stepanov went as a far as to suggest that the constructors could be free functions http://www.stlport.org/resources/StepanovUSA.html.
There is a way around it, which is to systematically declare a template<class Other> A& operator=(Other const& t); in inside all your classes, in this way you leave the option to anyone to define a custom assignment operator.
Of course you can't do this with a class you don't have control over.
Having said that it is nowadays not that bad since we have move assignment. And in some sense conversion operators B::operator A() const{...} are almost like a custom copy assignment. Conversion operators are now usable because of explicit. However you have to have control over the second type (B), the right-hand type in assignment.
Next question is why conversion operator need to be members them selves? Again, I don't think there is a good reason.
Related
I don't understand why in the following code the expression C c3 = 5 + c;
doesn't get compiled although 5 could be converted to type C as in the previous statement.
#include <iostream>
class C
{
int m_value;
public:
C(int value): m_value(value) {} ;
int get_value() { return m_value; } ;
C operator+(C rhs) { return C(rhs.m_value+m_value); }
};
int main()
{
C c = 10;
C c2 = c + 5; // Works fine. 5 is converted to type C and the operator + is called
C c3 = 5 + c; // Not working: compiler error. Question: Why is 5 not converted to type C??
std::cout << c.get_value() << std::endl; // output 10
std::cout << c2.get_value() << std::endl; // output 15
}
Because if overload operator as member function of the class, it could only be called when the object of that class is used as left operand. (And the left operand becomes the implicit *this object for the member function to be called.)
Binary operators are typically implemented as non-members to maintain symmetry (for example, when adding a complex number and an integer, if operator+ is a member function of the complex type, then only complex+integer would compile, and not integer+complex).
From the standard, [over.match.oper]/3
(emphasis mine)
For a unary operator # with an operand of a type whose cv-unqualified
version is T1, and for a binary operator # with a left operand of a
type whose cv-unqualified version is T1 and a right operand of a type
whose cv-unqualified version is T2, four sets of candidate functions,
designated member candidates, non-member candidates, built-in
candidates, and rewritten candidates, are constructed as follows:
(3.1) If T1 is a complete class type or a class currently being defined, the set of member candidates is the result of the qualified
lookup of T1::operator# ([over.call.func]); otherwise, the set of
member candidates is empty.
That means if the type of left operand is not a class type, the set of member candidates is empty; the overloaded operator (as member function) won't be considered.
You can overload it as a non-member function to allow the implicit conversion for both left and right operands.
C operator+(C lhs, C rhs) { return C(lhs.get_value() + rhs.get_value()); }
then both c + 5 or 5 + c would work fine.
LIVE
BTW: This will cause one temporaray object being constructed (from int to C) for the non-member function to be called; if you care about that, you can add all the three possible overloads as follows. Also note that this is a trade-off issue.
C operator+(C lhs, C rhs) { return C(lhs.get_value() + rhs.get_value()); }
C operator+(C lhs, int rhs) { return C(lhs.get_value() + rhs); }
C operator+(int lhs, C rhs) { return C(lhs + rhs.get_value()); }
And here're some suggestions about when to use a normal, friend, or member function overload.
In most cases, the language leaves it up to you to determine whether
you want to use the normal/friend or member function version of the
overload. However, one of the two is usually a better choice than the
other.
When dealing with binary operators that don’t modify the left operand
(e.g. operator+), the normal or friend function version is typically
preferred, because it works for all parameter types (even when the
left operand isn’t a class object, or is a class that is not
modifiable). The normal or friend function version has the added
benefit of “symmetry”, as all operands become explicit parameters
(instead of the left operand becoming *this and the right operand
becoming an explicit parameter).
When dealing with binary operators that do modify the left operand
(e.g. operator+=), the member function version is typically preferred.
In these cases, the leftmost operand will always be a class type, and
having the object being modified become the one pointed to by *this is
natural. Because the rightmost operand becomes an explicit parameter,
there’s no confusion over who is getting modified and who is getting
evaluated.
You are facing the reason to define certain operator overloads as free functions, i.e., when implicit conversions are desired. To see what's going on under the hood, consider the verbose form of operator overload invocations:
C c2 = c.operator+(5); // Ok, c has this member function
C c3 = 5.operator+(c); // No way, this is an integer without members
You can obviously do is an explicit C construction as in
C c3 = C{5} + c;
but this is not intended for an arithmetic value type like C. To make the implicit construction possible, define the overload as a free function
auto operator + (C lhs, const C& rhs)
{
lhs += rhs;
return lhs;
}
Now, there is no restriction of the left hand side operand. Note that the operator is implemented in terms of += (you would have to implement it to make the above code compile), which is good practice as pointed out in this thread: when you provide a binary operator + for a custom type, users of that type will also expected operator += to be available. Hence, to reduce code duplication, it's usually good to implement + in terms of += (same for all other arithmetic operands).
Further note that these operands often require a substantial amount of boilerplate code. To reduce this, consider e.g. the Boost operators library. To generate all standard arithmetic operators based on the minimal amount of actual hand-written code:
#include <boost/operators.hpp>
class C : private boost::arithmetic<C>
// ^^^^^^^^^^^^^^^^^^^^
// where the magic happens (Barton-Nackmann trick)
{
int m_value ;
public:
C(int value): m_value(value) {} ;
C& operator+=(const C& rhs) { m_value += rhs.m_value; return *this; }
C& operator-=(const C& rhs) { m_value -= rhs.m_value; return *this; }
C& operator*=(const C& rhs) { m_value *= rhs.m_value; return *this; }
C& operator/=(const C& rhs) { m_value /= rhs.m_value; return *this; }
const C& operator+() const { return *this; }
C operator-() const { return {-m_value}; }
int get_value() { return m_value; } ;
};
Here's an additional remark (a bit of a "reductio ad absurdum") on why your suggestion that the compiler could implicitly convert the left hand argument to a C would, essentially, open a can of worms. The actual language rules say, simply put, that, before applying conversions, a name lookup – for function calls and calls to (user-declared) operators – is done to find a candidate set. At this point, the operand types are not yet considered, but the scope very well is. So the type of the first argument does matter insofar as a user-declared operator is only in scope if its first argument is of the (cv-qualified) class type it is declared in. When a candidate set has been found, the compiler then tries to apply the conversion rules and ranks the candidates etc.
(Your question is therefore a bit misleading because in your example, we don't even get to the conversion logic, instead name resolution already comes up empty.)
Now, imagine we could simply change the language to say that the first argument can also be converted, prior to name resolution. A little bit of handwaving is required here, because this means we have to do conversions, look up names, and then do conversions again, so how this would work in practice is certainly unclear. Anyway, look at this example then:
struct B;
struct A
{
A(int);
A operator +(B) const;
};
struct B
{
B(int);
B operator +(B) const;
};
Now, what should 1 + B{3} do? Apparently, it could be transformed to B{1} + B{3}. But who's to say we couldn't do A{1} + B{3} instead? Why would B's constructor be preferred over A's? Of course, we could argue that either B is to be preferred, because, look at how nice and symmetric B{...}+B{...} is (ok, I'm being slightly facetious). Or we could take the safer route of saying that the program is ill-formed if it contains such an ambiguity. But there are a lot more corner cases to consider, e.g. what if B's constructor was made explicit – should the compiler (still or newly) error out, or should it silently switch to the usable implicit conversion to A?
Another non-obvious point is which types in which scopes (e.g. namespaces) should be considered? It would certainly be surprising if you use operator + in e.g. global namespace scope, and the compiler would dig out some type __gnucxx::__internal::__cogwheels::__do_something_impl, implcitly convert an operand to it, and then perform an operation on that.
Also note that this feature even if it can be specified in a reasonable and clean manner, could have quite a compile-time cost (in fact, overload resolution is already one of the biggests costs when compiling C++ and one of the reasons why compiling C++ code can take a lot longer than compiling C).
TL;DR:
There are tricky corner cases.
The benefit is marginal (why not make such operators free functions as others have pointed out)?
The discussions on how to standardize this would certainly be long.
I've read the section 13.5 of the working draft N3797 and I've one question. Let complex be a class type which represents complex numbers. We can define the following operator function:
complex operator+(double d, complex c)
{
return *new complex(d+c.get_real(),c.get_imagine());
}
But how this operator function can be implements as a complex member function? Or must I to declare this operator function in every module which I've intend to use them?
There are two ways to define a binary operator overload, such as the binary + operator, between two types.
As a member function.
When you define it as member function, the LHS of the operator is an instance of the class. The RHS of the operator is the argument to the function. That's why when you define it as member function, it can only have one argument.
As a free function.
These functions must have two arguments. The first argument is the LHS of the operator and the second argument is the RHS of the operator.
Since double is not a class, you have to define operator+ overload between double as LHS and complex as RHS as a free function, with double const& or double as the first argument type and complex const& or complex as the second argument type.
What you're looking for is this:
inline complex operator +(double d, const complex& c)
{
return complex(d+c.get_real(), c.get_imagine());
}
This cannot be a member function if you want the operator to handle a left-side of the operator as a double. It must be a free function, possibly friended if access to anything besides the public interface of complex is needed.
If you want the right side of the add-op to be a double, a member function can be crafted:
complex operator +(double d) const
{
return complex(d+get_real(), get_imagine());
}
Note this is assuming this definition is within the body of the complex class definition. But in the interest of clarity I would recommend both be inline free functions.
Implicit Construction
Lined up with the usual suspects, what you're at-least-appearing to try to do is generally done with an implicit conversion constructor and a general free-function. By providing this:
complex(double d) : real(d), imagine()
{
}
in the class definition a double can implicitly construct a temporary complex where needed. This allows this:
inline complex operator +(const complex& lhs, const complex& rhs)
{
return complex(lhs.get_real() + rhs.get_real(),
lhs.get_imagine() + rhs.get_imagine());
}
to be used as a general solution to all sensible manifestations of what you appear to want.
In C++, operator overloading requires class types. You cannot overload the + operator for the basic type double.
Also, except in certain circumstances (short-lived programs or throwaway code), the pointer resulting from a call to the new operator should be captured, so that it can be later released with delete, preventing a situation known as a memory leak.
For it to be a member, the first parameter must be of the class type(though it is not explicitly specified). But you are sending a double as first argument. If you need this to work either make it friend function or just non-member function which can work only using public interface of the Complex class. And as others pointed out, you are leaking memory.
If I want to overload operator +, which prototype is correct?
D operator+(const D& lhs, const D& rhs);
then declare it as a friend function of D.
D operator+(const D& s);
Then declare it as a member function of D.
The first is correct, the second is plain wrong. You can improve the second by writing this
D operator+(const D& s) const;
but it's still wrong. The reason is that the compiler will apply different rules to the left and right hand sides of your + operator in the second version. For instance given this code
class C
{
};
class D
{
public:
D(const C&);
};
C c;
D d;
d = d + c; // legal with both versions
d = c + d; // not legal with the second version
The difference is because the compiler will create a temporary D object from a C object for a method or function argument but it won't do it to make a method call on the temporary object.
In short the first version treats the left hand side and right hand side equally and so agrees better with the coders expectations.
The second one should be
D operator+(const D& s) const;
Then either is good.
As to the first needing to be a friend: only if it really needs access to anything private. Normally you can implement it in terms of the public interface, commonly with the corresponding operator+=:
D operator+(D lhs, const D& rhs) { //copy left-hand side
return lhs += rhs; //add the right-hand side and return by value
}
I would advice you to follow a third path: Implement operator+= as a member function, and then implement operator+ in terms of the previous like:
D operator+=( D lhs, D const & rhs ) {
lhs += rhs;
return lhs;
}
The advantage of the third way is that with basically the same code you provide both + and +=, and you get to implement operator+ as a free function which is an advantage from the point of view of symmetry, if your class has implicit conversions, it will allow d + t and t + d for any object d of type D and any other object t of type implicitly convertible to D. The member function version will only apply conversions to the right hand side, which means that d + t will be allowed, but not t + d.
[self publicity warning] You can read a longer explanation on this particular issue here
Go with the first one. However, if it needs to access private members,only then make it friend, otherwise make it non-friend function.
I think both are correct. But one thing you missed (that may or may not apply), is that if the left side value can be something other than D (say an integer or something) then option 1 works for that e.g.
D operator+(int lhs, const D& rhs);
Then you can do something like:
D d1;
D d2 = 5 + d1;
You only need to declare a function as a friend of a class if you plan to access private variables of the class through this function. In the first prototype you do not need to declare the function as a friend since you are passing in all the values you plan to use. In the second prototype you can declare the function a friend since you will need to access one more variable, but it is better practice and easier to just declare the function as a member.
Both methods are almost correct. It's just two ways of doing almost the same. But when you need to apply the binary operator to other types other than D (for example int+D) you need to use the second one.
The option 1 does not even have to be a friend if it does not need access to the private members.
The option 2 have to be fixed a little. You are missing D::.
D D::operator+(const D& s) const {
}
The principle of least surprise says that your overload should behave
more or less like the built-in operators. The usual solution here is
not to implement operator+ at all, but to implement:
D& operator+=( D const& rhs );
as a member, and then derive from something like:
template<typename T>
class ArithmeticOperators
{
friend T operator+( T const& lhs, T const& rhs )
{
T result( lhs );
result += rhs;
return result;
}
// Same thing for all of the other binary operators...
};
This way, you don't have to rewrite the same thing every time you define
a class which overloads the arithmetic operators, and you're guaranteed
that the semantics of + and += correspond.
(The friend in the above is simply to allow you to put the function,
along with its implementation, in the class itself, where ADL will find
it.)
Both are correct (after fixing the const correctness problem others pointed out), but I recommend the member operator. It can be overridden with polymorphic behavior if virtual, have better cohesion with the class and as a result it is easier to maintain and to document. Try to avoid friend functions if you can.
As for the "derived = base + derived" case, I recommend not mixing value semantics operators and polymorphism, it can have unforeseen consequences due to various implicit conversion sequences and object slicing. That example might be equivalent to derived = Derived(base) + derived, but it can be derived = Derived(base + Base(derived)) as well if base has operator +.
Use only explicit conversion and casting, and you will not encounter any mysterious strange behavior. And think twice before you implement operators for a polymorphic class.
The first one has a different behavior if D has implicit constructors.
Consider
struct D
{
D(int);
};
D operator+(const D&, const D&);
Then you can do 1 + d and d + 1, which you cannot do with the member operator+ (the lhs must be a D, and no conversion shall take place).
Hi I have a code like this, I think both the friend overloaded operator and conversion operator have the similar function. However, why does the friend overloaded operator is called in this case? What's the rules?
Thanks so much!
class A{
double i;
public:
A(int i):i(i) {}
operator double () const { cout<<"conversion operator"<<endl;return i;} // a conversion operator
friend bool operator>(int i, A a); // a friend funcion of operator >
};
bool operator>(int i, A a ){
cout<<"Friend"<<endl;
return i>a.i;
}
int main()
{
A aa(1);
if (0 > aa){
return 1;
}
}
No conversion is necessary for the overloaded operator> to be called. In order for the built-in operator> to be called, one conversion is necessary (the user-defined conversion operator. Overload resolution prefers options with fewer required conversions, so the overloaded operator> is used.
Note that if you were to change the definition of your overloaded operator> to be, for example:
friend bool operator>(double i, A a);
you would get a compilation error because both the overloaded operator> and the built-in operator> would require one conversion, and the compiler would not be able to resolve the ambiguity.
I am not claiming that my answer is supported by the standards, but lets think about it logically.
When you hit this line:
0 > aa
You have two options. Either you call the provided operator:
friend bool operator>(int i, A a);
Which is 100% compatible, or you can do two conversions to reach your destination! Which one would you choose?
If you add a conversion operator then an object of type A may be converted to double when you least expect it.
A good program does not provide a way for his classes to be accidently used and the conversion operator opens up the opertunity for the class to be used in a whole host of unintended situations (normally situations where you would expect a compile time error are not because of the auto type conversion).
As a result conversion operators (and single argument constructors) should be treateed with some care because of situations were the compiler may do conversion when you least expect it.
It is called because it is an exact match in the context of the expression 0 > aa. In fact, it is hard to figure out how you came up with the "why" question. By logic, one'd expect a "why" question if the friend weren't called in this case.
If you change the expression to 0.0 > aa, the call will beciome ambiguous, because neuther path will be better than the other.
I wrote an abstraction class for a math object, and defined all of the operators. While using it, I came across:
Fixed f1 = 5.0f - f3;
I have only two subtraction operators defined:
inline const Fixed operator - () const;
inline const Fixed operator - (float f) const;
I get what is wrong here - addition is swappable (1 + 2 == 2 + 1) while subtraction is not (same goes for multiplication and division).
I immediately wrote a function outside my class like this:
static inline const Fixed operator - (float f, const Fixed &fp);
But then I realized this cannot be done, because to do that I would have to touch the class's privates, which results to using the keyword friend which I loath, as well as polluting the namespace with a 'static' unnecessary function.
Moving the function inside the class definition yields this error in gcc-4.3:
error: ‘static const Fixed Fixed::operator-(float, const Fixed&)’ must be either a non-static member function or a non-member function
Doing as GCC suggested, and making it a non-static function results the following error:
error: ‘const Fixed Fixed::operator-(float, const Fixed&)’ must take either zero or one argument
Why can't I define the same operator inside the class definition? if there's no way to do it, is there anyway else not using the friend keyword?
Same question goes for division, as it suffers from the same problem.
If you need reassuring that friend functions can be OK:
http://www.gotw.ca/gotw/084.htm
Which operations need access to
internal data we would otherwise have
to grant via friendship? These should
normally be members. (There are some
rare exceptions such as operations
needing conversions on their left-hand
arguments and some like operator<<()
whose signatures don't allow the *this
reference to be their first
parameters; even these can normally be
nonfriends implemented in terms of
(possibly virtual) members, but
sometimes doing that is merely an
exercise in contortionism and they're
best and naturally expressed as
friends.)
You are in the "operations needing conversions on the left-hand arguments" camp. If you don't want a friend, and assuming you have a non-explicit float constructor for Fixed, you can implement it as:
static inline Fixed operator-(const Fixed &lhs, const Fixed &rhs) {
return lhs.minus(rhs);
}
then implement minus as a public member function, that most users won't bother with because they prefer the operator.
I assume if you have operator-(float) then you have operator+(float), so if you don't have the conversion operator, you could go with:
static inline Fixed operator-(float lhs, const Fixed &rhs) {
return (-rhs) + lhs;
// return (-rhs) -(-lhs); if no operator+...
}
Or just Fixed(lhs) - rhs if you have an explicit float constructor. Those may or may not be as efficient as your friend implementation.
Unfortunately the language is not going to bend over backwards to accommodate those who happen to loathe one of its keywords, so operators can't be static member functions and get the effects of friendship that way ;-p
"That's what friends are for..."
You could add an implicit conversion between float and your type (e.g. with a constructor accepting float)... but I do think using a friend is better.
When you define something like this,
inline const Fixed operator - (float f) const;
you are saying that I want this operator(you are inside the class) to operate on a specific type, float here for example.
Whereas a friend binary operator, means an operation between two types.
class Fixed
{
inline friend const Fixed operator-(const Fixed& first, const float& second);
};
inline const Fixed operator-(const Fixed& first, const float& second)
{
// Your definition here.
}
with friend operators you can have your class on either side of the operator it self.
In general, free function operators for arithmetic operations are better than implementing member functions. The main reason is the problem you are facing now. The compiler will treat the left and right sides differently. Note that while strict OO followers will consider only those methods inside the class curly braces part of its interface, but it has been argued by experts that not to be the case in C++.
If the free function operator requires access to private members, make the operator friend. After all if it is provided in the same header file (following Sutter's rationale above) then it is part of the class.
If you really want to avoid it and don't mind making your code less idiomatic (and thus less maintainable) you can provide a public method that does the real work and dispatch to that method from the operator.
class Fixed {
private:
Fixed();
Fixed( double d ); // implicit conversion to Fixed from double
Fixed substract( Fixed const & rhs ) const;
// ...
};
Fixed operator-( Fixed const & lhs, Fixed const & rhs )
{
return lhs.substract( rhs );
}
In the code above, you can substract Fixed - Fixed, Fixed - double, double - Fixed. The compiler will find the free function and implicitly convert (in the example through the double constructor) the doubles into Fixed objects.
While this is unidiomatic for arithmetic operators, it is close to the idiomatic way of proving a polymorphic dump operator. So while not being the most natural solution it won't be the most surprising code around either
// idiomatic polymorphic dump operator
class Base {
public:
virtual std::ostream& dump( std::ostream & ) const;
};
std::ostream& operator<<( std::ostream& o, Base const & d )
{
return d.dump( o );
}