Why is it giving the error.
class Complex{
int real,imaginary;
public:
Complex():real(0),imaginary(0){};
Complex(int real, int imaginary):real(real),imaginary(imaginary){};
int getreal()const{return real;}
int getimaginary()const{return imaginary;};
Complex &operator=(const Complex& );
Complex operator*(){
return Complex(real,-imaginary);
}
Complex operator+(const Complex &a, const Complex &d){
return Complex (d.getreal()+a.getreal(),a.getimaginary()+d.getimaginary());
}
};
When trying to overload the assingment operator.
Eroor says: too many parameter
What operators can we overload as member function and what not
When overriding operators that have left and right parameters, you should just pass the right parameter and declare it like this:
Complex operator+(const Complex &d){
return Complex(d.getreal() + getreal(), getimaginary() + d.getimaginary());
}
When doing that, when you call getreal() or directly access that variable without specifing the parameter, it will use the left parameter of the operator.
Some of the operators that cannot be overloaded are scope (::), ternary (:), sizeof, member access (.) and some others.
When defining an overloaded operator as a member function, the object pointed at by the this pointer is implicitly the first argument. So your operator overload needs to look like:
class Complex {
// ...
Complex operator+(const Complex &d) const {
return Complex (real + d.real, imaginary + d.imaginary);
}
};
Note that you don't need the getter functions, since you already have access to the data members inside the class.
Outside the class, i.e. as a non-member function, your overloaded operator is perfectly fine.
Related
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.
I want to overload the * operator for a class, one which is in case it is multiplied by a real number double and the other is when multiplied by the same class type.
Below is my code:
class Vector2D {
...
friend Vector2D operator* (const Vector2D&);
friend Vector2D operator* (const double); // getting error here
};
The error I'm getting is: 'Vector2D operator*(double)' must have an argument of class or enumerated type
Is it possible to overload operators more than once, and if so, why am I getting an error?
class Vector2D {
...
friend Vector2D operator* (const Vector2D&);
That is overloading unary operator* (i.e. dereference operator) as a free function, not overloading the multiplication of two vectors. You cannot overload operators that apply only to non-class types, and that is the error for:
friend Vector2D operator* (const double);
You probably meant to write:
friend Vector2D operator*(const Vector2D&,const Vector2D&);
friend Vector2D operator*(const Vector2D&,double);
The problem is that you're overloading unary operator*, not binary. Your friend declarations are not member functions.
Either remove the friend (if you want to overload the operators as member functions), or else add a first parameter of type const Vector2D& or Vector2D& (if you want to overload as free functions). The latter is usually preferable, since then you get implicit conversion on both arguments -- member operator overloads will implicitly convert the right hand side but not the left.
The reason you might not want a reference as the first argument to binary operator*, is so that you can implement it like this:
Vector2D operator*(Vector2D lhs, const Vector2D &rhs) {
lhs *= rhs;
return lhs;
}
The same would apply to addition (and addition of two vectors makes sense to me in a way that multiplication of two vectors does not).
You are not allowed to override the operators for built-in types. In this case you're providing a declaration for a unary * operator for type double which isn't allowed.
I think you want to remove the friend declaration and make these member functions instead.
// By using structure :
struct complex {
float real;
float imag;
};
complex operator+(complex, complex);
main() {
complex t1, t2, t3;
t3 = t1 + t2;
}
complex operator+(complex w, complex z) {
statement 1;
statement 2;
}
// By using class :
class complex {
int real;
int imag;
public:
complex operator+(complex c) {
statement 1;
statement 2;
}
main() {
complex t1, t2, t3;
t3 = t1 + t2;
}
While using structure, the overloaded function can accept two arguments whereas while using class the overloaded function accepts only one argument, when the overloaded operator function is a member function in both cases i.e in struct as well as in class. Why does this happen?
That has nothing to do with classes vs. structs. It's about member vs. nonmember.
Classes and structs in C++ differ solely bu their default accessibility level for members and bases (public for structs, and private for classes). Other than this, there is no difference at all.
When overloading operators, you almost always have the choice of defining an operator as a member or as a freestanding function. There are only 4 operators that have to be members. These are: (), [], ->, and = (as to why, see this question of mine). For the rest, the choice is yours.
This excellent FAQ entry explains (among other things) how to choose between member vs. nonmember.
To answer your core question: In case of member function, the first agument is *this
...when the overloaded operator function is a member function in both
cases i.e structure as well as class...
What makes you say that? That's not true.
In case of struct in your example, overloaded operator function is not a member. This is why it requires 2 parameters.
So, the difference has absolutely nothing to do with struct vs. class matter. The reason you have different number of parameters in these operators is that the first one is implemented as a non-member (and therefore has two explicit parameters), while the second one is implemented as member (and therefore has only one explicit parameter).
This happens because you using different methods to define operator. When you use struct you define it outside of struct, but when use class you define operator inside class.
When you want to pass two args to your overloaded operator+ you should create a friend function of the class and then pass const complex& lhs, const complex& rhs as your args. Now you can have two arguments to your operator+.
When the operator+ is a member of the class, one arg is implicitly passed as the this pointer.
//
complex operator +(cont complex& lhs, const complex& rhs)
{
...
}
main()
{
complex t1,t2,t3;
t3=t1+t2; // i.e. t3 = operator+(t1, t2);
}
//
class complex
{
int real,imag;
public:
complex operator +(const complex& rhs){}
};
main()
{
complex t1,t2,t3;
t3=t1+t2; // i.e. t3 = t1.operator+(t2);
}
In your second example the overloaded operator is a member function of the class, therefore when this operator is called, it's called as a member function of the first operand and the first operand is therefore implicitly given as the object on which the member function has been called (the this pointer if you like).
I read that an overloaded operator declared as member function is asymmetric because it can have only one parameter and the other parameter passed automatically is the this pointer. So no standard exists to compare them. On the other hand, overloaded operator declared as a friend is symmetric because we pass two arguments of the same type and hence, they can be compared.
My question is that when i can still compare a pointer's lvalue to a reference, why are friends preferred? (using an asymmetric version gives the same results as symmetric)
Why do STL algorithms use only symmetric versions?
If you define your operator overloaded function as member function, then the compiler translates expressions like s1 + s2 into s1.operator+(s2). That means, the operator overloaded member function gets invoked on the first operand. That is how member functions work!
But what if the first operand is not a class? There's a major problem if we want to overload an operator where the first operand is not a class type, rather say double. So you cannot write like this 10.0 + s2. However, you can write operator overloaded member function for expressions like s1 + 10.0.
To solve this ordering problem, we define operator overloaded function as friend IF it needs to access private members. Make it friend ONLY when it needs to access private members. Otherwise simply make it non-friend non-member function to improve encapsulation!
class Sample
{
public:
Sample operator + (const Sample& op2); //works with s1 + s2
Sample operator + (double op2); //works with s1 + 10.0
//Make it `friend` only when it needs to access private members.
//Otherwise simply make it **non-friend non-member** function.
friend Sample operator + (double op1, const Sample& op2); //works with 10.0 + s2
}
Read these :
A slight problem of ordering in operands
How Non-Member Functions Improve Encapsulation
It's not necessarily a distinction between friend operator overloads and member function operator overloads as it is between global operator overloads and member function operator overloads.
One reason to prefer a global operator overload is if you want to allow expressions where the class type appears on the right hand side of a binary operator. For example:
Foo f = 100;
int x = 10;
cout << x + f;
This only works if there is a global operator overload for
Foo operator + (int x, const Foo& f);
Note that the global operator overload doesn't necessarily need to be a friend function. This is only necessary if it needs access to private members of Foo, but that is not always the case.
Regardless, if Foo only had a member function operator overload, like:
class Foo
{
...
Foo operator + (int x);
...
};
...then we would only be able to have expressions where a Foo instance appears on the left of the plus operator.
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 );
}