Suppose I have the following class:
class Point{
private:
int x,y;
public:
int get_x() const {return x;}
int get_y() const {return y;}
Point() :x(0),y(0){}
Point(int x,int y):x(x),y(y){}
Point(const Point& P){
x = P.get_x();
y = P.get_y();
}
Point& operator= (const Point& P) {
x = P.get_x();
y = P.get_y();
return *this;
}
friend ostream& operator<<(ostream& os,const Point& P) {
os<<"["<<P.get_x()<<", "<<P.get_y()<<"]";
return os;
}
Point operator - (const Point &P){
return Point(x-P.get_x(),y-P.get_y());
}
friend bool operator > (const Point &A, const Point &B) {
return A.get_y()>B.get_y();
}
};
Here I used friend function. I can also use function without friend:
class Point{
...
bool operator > (const Point &B) const {
return y>B.get_y();
}
...
};
What are the differences between them in actual implementations? Also in the second method, the code won't compile without 'cont', why is that? Even after I changed the getter function into non-const function, it still won't compile without the 'const'.
As you've already noticed, comparison operator overloads can either be implemented as a member function or as a non-member function.
As a rule of thumb you should implement them as a non-member non-friend function where possible, as this increases encapsulation, and it allows (non-explicit) conversion constructors to be used on either side of the operator.
Say for instance your Point class for whatever reason had an int conversion constructor:
Point(int x);
With a non-member comparison operator you can now do the following:
Point p;
p < 3; // this will work with both a member and non-member comparison
3 < p; // this will **only** work if the comparison is a non-member function
You also seem to be confused about when to use const, again as a rule of thumb for comparison operators you should always use const wherever possible, because comparisons logically do not involve any change to the object.
As Point is a very small class you could also take it by value instead, so in order of most to least preferable your options are:
// Non-member, non-friend
bool operator>(Point const& A, Point const& B);
bool operator>(Point A, Point B);
// Non-member, friend
friend bool operator>(Point const& A, Point const& B);
friend bool operator>(Point A, Point B);
// Member
bool Point::operator>(Point const& B) const;
bool Point::operator>(Point B) const;
Related
Is there a difference between defining a global operator that takes two references for a class and defining a member operator that takes only the right operand?
Global:
class X
{
public:
int value;
};
bool operator==(X& left, X& right)
{
return left.value == right.value;
};
Member:
class X
{
int value;
bool operator==( X& right)
{
return value == right.value;
};
}
One reason to use non-member operators (typically declared as friends) is because the left-hand side is the one that does the operation. Obj::operator+ is fine for:
obj + 2
but for:
2 + obj
it won't work. For this, you need something like:
class Obj
{
friend Obj operator+(const Obj& lhs, int i);
friend Obj operator+(int i, const Obj& rhs);
};
Obj operator+(const Obj& lhs, int i) { ... }
Obj operator+(int i, const Obj& rhs) { ... }
Your smartest option is to make it a friend function.
As JaredPar mentions, the global implementation cannot access protected and private class members, but there's a problem with the member function too.
C++ will allow implicit conversions of function parameters, but not an implicit conversion of this.
If types exist that can be converted to your X class:
class Y
{
public:
operator X(); // Y objects may be converted to X
};
X x1, x2;
Y y1, y2;
Only some of the following expressions will compile with a member function.
x1 == x2; // Compiles with both implementations
x1 == y1; // Compiles with both implementations
y1 == x1; // ERROR! Member function can't convert this to type X
y1 == y2; // ERROR! Member function can't convert this to type X
The solution, to get the best of both worlds, is to implement this as a friend:
class X
{
int value;
public:
friend bool operator==( X& left, X& right )
{
return left.value == right.value;
};
};
To sum up to the answer by Codebender:
Member operators are not symmetric. The compiler cannot perform the same number of operations with the left and right hand side operators.
struct Example
{
Example( int value = 0 ) : value( value ) {}
int value;
Example operator+( Example const & rhs ); // option 1
};
Example operator+( Example const & lhs, Example const & rhs ); // option 2
int main()
{
Example a( 10 );
Example b = 10 + a;
}
In the code above will fail to compile if the operator is a member function while it will work as expected if the operator is a free function.
In general a common pattern is implementing the operators that must be member functions as members and the rest as free functions that delegate on the member operators:
class X
{
public:
X& operator+=( X const & rhs );
};
X operator+( X lhs, X const & rhs )
{
lhs += rhs; // lhs was passed by value so it is a copy
return lhs;
}
There is at least one difference. A member operator is subject to access modifiers and can be public, protected or private. A global member variable is not subject to access modifier restrictions.
This is particularly helpful when you want to disable certain operators like assignment
class Foo {
...
private:
Foo& operator=(const Foo&);
};
You could achieve the same effect by having a declared only global operator. But it would result in a link error vs. a compile error (nipick: yes it would result in a link error within Foo)
Here's a real example where the difference isn't obvious:
class Base
{
public:
bool operator==( const Base& other ) const
{
return true;
}
};
class Derived : public Base
{
public:
bool operator==( const Derived& other ) const
{
return true;
}
};
Base() == Derived(); // works
Derived() == Base(); // error
This is because the first form uses equality operator from base class, which can convert its right hand side to Base. But the derived class equality operator can't do the opposite, hence the error.
If the operator for the base class was declared as a global function instead, both examples would work (not having an equality operator in derived class would also fix the issue, but sometimes it is needed).
Consider a class SomeClass:
class SomeClass{
public:
// Constructors, other members
float& operator[](const unsigned i);
friend bool operator==(const SomeClass &A, const SomeClass &B);
};
Suppose this is how the == operator is overloaded for this class (not the actual implementation, but an overly simplified version):
bool operator==(const SomeClass &A, const SomeClass &B){
if (A[0] == B[0])
return true;
return false;
}
This would throw a compiler error, since the overloaded [] operator requires the instance to be non-const. However, if I change the definition of the [] operator to allow for const instances, I can no longer do assignment:
// ASSUMING: const float& operator[](const unsigned i) const;
SomeClass a;
a[0] = 0; // error, because the return value of [] is a reference to const!
I really don't want to drop the const in the parameters of the == operator, since the operands don't change within the function. What is the right way to deal with this issue?
Overload operator [] to provide both:
float& operator [](unsigned int i);
float operator [](unsigned int i) const;
For a generic T that’s not cheap to copy, use a T const& return value. The general pattern for implementing read/write operator [] is
T& operator [](index_type i);
T const& operator [](index_type i) const;
I am stuck with regards to the 2 non-member, 2 non-friend multiplication and the addition of the operator overloading functions. I am unsure of how to do it. Could someone please assist in helping me to resolve this? Refer to my codes below. Thank you in advance!
Compiler output:
Point.cpp:208:19: error: passing ‘const CS170::Point’ as ‘this’ argument discards qualifiers [-fpermissive]
return other + value;
^~~~~
Point.cpp: In function ‘CS170::Point CS170::
operator*(double, const CS170::Point&)’:
Point.cpp:215:10: error: ‘double CS170::Point::x’ is private within this context
result.x = value * x;
^
Point.cpp:215:22: error: ‘x’ was not declared in this scope
result.x = value * x;
^
Point.cpp:216:10: error: ‘double CS170::Point::y’ is private within this context
result.y = value * y;
^
Point.cpp:216:23: error: ‘y’ was not declared in this scope
result.y = value * y;
Point.h
#include <iostream> // istream, ostream
namespace CS1100
{
class Point
{
public:
// Point(double X, double Y); // Constructors (2)
explicit Point(double x, double y);
Point();
Point operator+(const Point& other)const ;
Point& operator+(double value);
Point operator*(double value) ;
Point operator%(double value);
Point operator-(const Point& other)const ;
Point operator-(double value);
Point operator^(const Point& other);
Point operator+=(double value);
Point& operator+=(const Point& other) ;
Point& operator++();
Point operator++(int);
Point& operator--();
Point operator--(int);
Point& operator-();
// Overloaded operators (14 member functions)
friend std::ostream &operator<<( std::ostream &output, const Point &point );
friend std::istream &operator>>( std::istream &input, Point &point );
// Overloaded operators (2 friend functions)
private:
double x; // The x-coordinate of a Point
double y; // The y-coordinate of a Point
// Helper functions
double DegreesToRadians(double degrees) const;
double RadiansToDegrees(double radians) const;
};
// Point& Add(const Point& other); // Overloaded operators (2 non-member, non-friend functions)
// Point& Multiply(const Point& other);
Point operator+( double value, const Point& other );
Point operator-( double value, const Point& other );
My source code:
///////////////////////////////////////////////////////////////////////////////
// 2 non-members, non-friends (operators)
double operator+( double value, const Point& other )
{
return other + value;
}
double operator*( double value, const Point& other )
{
Point result;
result.x = value * x;
result.y = value * y;
return result;
}
As far as I understand the discussion to the question, the problem is not really the operators themselves, but the number of allowed member functions being limited – and you already have exceeded this limit.
However, you have quite a number of functions that don't need to be members, for instance:
class Point
{
public:
Point operator+(const Point& other) const
{
return Point(x + other.x, y + other.y);
}
};
Make free functions from all these:
class Point { /*...*/ };
Point operator+(Point const& l, Point const& r)
{
return Point(l.getX() + r.getX(), l.getY() + r.getY());
}
Having moved out all these operators like the one shown above, you get away far enough from the limit so that you can introduce the needed getters:
class Point
{
public:
double getX() { return x; };
double getY() { return y; };
};
If you are willing to rename the member variables, e. g. by adding a prefix, you can follow another pattern:
class Point
{
double m_x, m_y;
public:
double x() { return m_x; };
double y() { return m_y; };
void x(double v) { m_x = v; }; // the corresponding setter
// (for illustration, you might not need it)
};
This latter pattern is quite common, too. Advantage is being shorter for skipping the explicit get or set prefix, disadvantage is exactly losing this explicitness... Decide you, which one you prefer. More important than personal preference is consistency, though, so if there's e. g. a company's convention or common practice, follow that one...
Some of your operators will need to remain members, though, these are all those that modify the current object:
class Point
{
public:
Point& operator+=(const Point& other) /* const */ // NEEDS to be non-const
{
x += other.x;
y += other.y;
return *this; // <- very good hint to spot the ones needing to stay members
}
};
If you have a public copy constructor, you can re-use the operator+= for defining the operator+:
class Point
{
public:
Point(Point const& other) : Point(other.x, other.y) { }
};
Point operator+(Point const& x, Point const& y)
{
Point r(x); // or Point(x.x(), x.y()), if you lack such constructor)
r += y;
return r;
}
Actually, you can even spare the explicit copy by accepting one of the parameters by value:
Point operator+(Point x, Point const& y)
// ^ no reference
{
return x += y;
}
The latter rather for illustration, I'd prefer the two references in given case to keep the symmetry of the interface...
So, i have a simple class:
class complex{
private:
double a,b;
public:
void setA(double a){ this->a=a; }
void setB(double b){ this->b=b; }
double getA(){ return a; }
double getB(){ return b; }
friend complex operator+(const complex&, const complex&);
};
And i have the actual overloaded operator here:
complex operator+(const complex& x, const complex& y){
complex c;
c.a=x.a+y.a;
c.b=x.b+y.b;
return c;
}
I must have the operator overloaded outside of the function. In order to have access to private variables (those also HAVE to be private) I befriended the class with the function. I don't know if that's correct way to do such things, but at least it works.
I want to be able to add an integer to both members. In main():
complex a;
a.setA(2);
a.setB(3);
a+5;
Would result in having a.a=7 and a.b=8. Such overload inside the class is quite easy to make (Again, don't know if that's good solution, if not please correct me):
complex operator+(int x){
this->a+=x;
this->b+=x;
}
But I have no idea how to make it outside of the class because i can't use "this" pointer.
The usual approach to this sort of problem is to have member functions that define the reflexive version of arithmetic operators and free functions that define the non-reflexive version, implemented with the reflexive version. No friend declarations needed. For example:
class complex {
public:
complex& operator+=(const complex& rhs) {
x += rhs.x;
y += rhs.y;
return *this;
}
private:
double x, y;
};
complex operator+(const complex& lhs, const complex& rhs) {
complex result = lhs;
result += rhs;
return result;
}
Having a+5 change the value of a is unusual, but if that's really wha you want, make operator+(int) a member. However, users would typically expect that a+5 would leave a unchanged, and that a += 5 would modify a.
Is there a difference between defining a global operator that takes two references for a class and defining a member operator that takes only the right operand?
Global:
class X
{
public:
int value;
};
bool operator==(X& left, X& right)
{
return left.value == right.value;
};
Member:
class X
{
int value;
bool operator==( X& right)
{
return value == right.value;
};
}
One reason to use non-member operators (typically declared as friends) is because the left-hand side is the one that does the operation. Obj::operator+ is fine for:
obj + 2
but for:
2 + obj
it won't work. For this, you need something like:
class Obj
{
friend Obj operator+(const Obj& lhs, int i);
friend Obj operator+(int i, const Obj& rhs);
};
Obj operator+(const Obj& lhs, int i) { ... }
Obj operator+(int i, const Obj& rhs) { ... }
Your smartest option is to make it a friend function.
As JaredPar mentions, the global implementation cannot access protected and private class members, but there's a problem with the member function too.
C++ will allow implicit conversions of function parameters, but not an implicit conversion of this.
If types exist that can be converted to your X class:
class Y
{
public:
operator X(); // Y objects may be converted to X
};
X x1, x2;
Y y1, y2;
Only some of the following expressions will compile with a member function.
x1 == x2; // Compiles with both implementations
x1 == y1; // Compiles with both implementations
y1 == x1; // ERROR! Member function can't convert this to type X
y1 == y2; // ERROR! Member function can't convert this to type X
The solution, to get the best of both worlds, is to implement this as a friend:
class X
{
int value;
public:
friend bool operator==( X& left, X& right )
{
return left.value == right.value;
};
};
To sum up to the answer by Codebender:
Member operators are not symmetric. The compiler cannot perform the same number of operations with the left and right hand side operators.
struct Example
{
Example( int value = 0 ) : value( value ) {}
int value;
Example operator+( Example const & rhs ); // option 1
};
Example operator+( Example const & lhs, Example const & rhs ); // option 2
int main()
{
Example a( 10 );
Example b = 10 + a;
}
In the code above will fail to compile if the operator is a member function while it will work as expected if the operator is a free function.
In general a common pattern is implementing the operators that must be member functions as members and the rest as free functions that delegate on the member operators:
class X
{
public:
X& operator+=( X const & rhs );
};
X operator+( X lhs, X const & rhs )
{
lhs += rhs; // lhs was passed by value so it is a copy
return lhs;
}
There is at least one difference. A member operator is subject to access modifiers and can be public, protected or private. A global member variable is not subject to access modifier restrictions.
This is particularly helpful when you want to disable certain operators like assignment
class Foo {
...
private:
Foo& operator=(const Foo&);
};
You could achieve the same effect by having a declared only global operator. But it would result in a link error vs. a compile error (nipick: yes it would result in a link error within Foo)
Here's a real example where the difference isn't obvious:
class Base
{
public:
bool operator==( const Base& other ) const
{
return true;
}
};
class Derived : public Base
{
public:
bool operator==( const Derived& other ) const
{
return true;
}
};
Base() == Derived(); // works
Derived() == Base(); // error
This is because the first form uses equality operator from base class, which can convert its right hand side to Base. But the derived class equality operator can't do the opposite, hence the error.
If the operator for the base class was declared as a global function instead, both examples would work (not having an equality operator in derived class would also fix the issue, but sometimes it is needed).