The other day, I ran into trouble when I tried to overload a two-parameter operator with the use of a class member function. I tried references, but nothing changed. The compiler said I couldn't write a member function that takes more than one argument of the same type as the class itself. Why is that?
Here is the code:
class Fraction
{
public:
Fraction(int num=1, int den=1): numerator(num), denominator(den) {}
Fraction(const Fraction& r): numerator(r.numerator), denominator(r.denominator) {}
Fraction& operator=(const Fraction&);
Fraction& operator*(const Fraction&, const Fraction&);
private:
int numerator, denominator;
};
Fraction& Fraction::operator=(const Fraction& r)
{
numerator = r.numerator;
denominator = r.denominator;
return *this;
}
Fraction Fraction::operator*(const Fraction& x, const Fraction& y)
{
Fraction z(x.numerator*y.numerator, x.denominator*y.denominator);
return z;
}
The following is the error message from the compiler:
Fraction& Fraction::operator*(const Fraction&, const Fraction&)' must take either zero or one argument
The operators take a fixed amount of arguments, for example operator+ as the addition operator takes exactly 2 parameters. If it is a member function the first(leftmost) is implied to be this and the second is passed as a parameter.
What would calling an operator+ that took three parameters look like? One might imagine it would look like 3 + 4 + 5 but that is the equivalent of calling operator+(3,4) then operator+(7,5).
For a list of operators and how many arguments they take please check out wikipedia: http://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B
For more details there is another StackOverflow post that goes into great detail: Operator overloading
Firstly, the issue you are observing has absolutely nothing to do with member functions in general. Member functions can generally take arbitrary number of arguments of "same class" type. In your example you can declare
class Fraction
{
void foo(Fraction &f1, Fraction &f2, Fraction &f3, Fraction &f4) {}
...
};
without any problems. So, it is not clear why you decided to word your question as if it is about member functions in general.
Secondly, in your code it is really about the simple fact that you are trying to overload an operator. Operator syntax in C++ is fixed for most (but not all) operators. This immediately means that those operators whose syntax is fixed will have fixed number of parameters.
In your example, it is operator *. It can be unary (one parameter) or binary (two parameters). When you overload this operator by a member function one parameter is already implied, so you can only add zero or one additional parameters (for unary * and binary * respectively). You, on the other hand, are trying to add two more parameters in additions to the implicit one. I.e. you are trying to define a ternary operator *. This is not possible. There's no ternary * operator in C++. And this is exactly what the compiler is telling you.
I'm trying to achieve this using parameter overloading using C++:
Complex c(3.0, 4.0);
double magnitude = | c; // magnitude will be 5
I wrote the following code: (Only the necessary part here..)
class Complex
{
public:
double _real;
double _imaginary;
friend double operator|(const Complex &c1)
{
return sqrt(c1._real * c1._real + c1._imaginary * c1._imaginary);
}
}
But I get the following error:
error C2805: binary 'operator |' has too few parameters
Is that impossible to use operator | with only 1 parameter?
friend double operator|(const Complex &c1)
{
return sqrt(c1._real * c1._real + c1._imaginary * c1._imaginary);
}
This doesn't define a member operator, just FYI.
double magnitude = | c;
this is invalid syntax, | is a binary operator.
Correct way:
class Complex
{
public:
double _real;
double _imaginary;
double getMagnitude() const // POP POP!
{
return sqrt(_real * _real + _imaginary * _imaginary);
}
}
No more bonus.
Is that impossible to use operator | with only 1 parameter?
You can overload operators as long as atleast one of the types involved is an user defined type but you cannot change the behavior w.r.t how many parameters they can take.
As the error message tells you | is a binary operator you cannot overload it to act as an unary operator.
What is the correct way to do this?
You should provide a utility function for your class Complex, do name it appropriately and it will to do the job for you in the best possible way.
Note that the very basic rule of operator overloading is:
"Whenever the meaning of an operator is not obviously clear and undisputed, it should not be overloaded. Instead, provide a function with a well-chosen name."
The rule is meant for non-intuitive operator usage like this.
operator| is a binary operator. As binary operator, it needs 2 parameters. If you want to do what you want to do here, you must use an unary operator.
Anyway - it looks like a bad idea since it is not obvious from looking at the operator to see what it does.
No - the '|' operator is a binary operator, meaning that it takes two parameters. You can overload operators but not change their "arity". Some operators are available with multiple arities though.
Unary operators include:
+
-
++ (both pre and post versions)
-- (both pre and post versions)
!
~
*
&
(cast) (but you would have to define a suitable casting type to get your double result)
The best solution from a software engineering point of view would probably be an explicit method to get the modulus - e.g. getModulus(). But you could legitimately argue that a double cast is OK.
For the latter case, you would have:
class Complex
{
public:
double _real;
double _imaginary;
operator double() const
{
return sqrt(this._real * this._real + this._imaginary * this._imaginary);
}
}
and use it as follows:
Complex c(3.0, 4.0);
double magnitude = c; // magnitude will be 5
Is that impossible to use operator | with only 1 parameter?
Yes. operator| is a binary operator. That means it takes two arguments. What you're looking for is operator|=
struct wtf
{
double operator|= (double omg)
{
return 42.;
}
};
int main(){ wtf omg; omg|= 42.; }
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Overload operators as member function or non-member (friend) function?
In process of learning operator overloading in C++ I have seen two different types of overloading operator +.
I need your help to tell me which method is better to use:
Method first:
Complex Complex::operator + (Complex &obj) {
return Complex( re + obj.re, im + obj.im );
}
Method second:
Complex operator + (const Complex &obj1, const Complex &obj2) {
// this function is friend of class complex
return Complex(obj1.re + obj2.re, obj1.im + obj2.im);
}
Thank you!!!
As long as there can be no implicit conversions, both are equally good;
it's more a matter of taste. If there can implicit conversions to
Complex (which in this case seems likely), then with the first form,
implicit conversions will only work on the second argument, e.g.:
Complex c;
Complex d;
d = c + 1.0; // Works in both cases...
d = 1.0 + c; // Only works if operator+ is free function
In such cases, the free function is by far the preferred solution; many
people prefer it systematically, for reasons of orthogonality.
In many such cases, in fact, the free function operator+ will be
implemented in terms of operator+= (which will be a member):
Complex
operator+( Complex const& lhs, Complex const& rhs )
{
Complex results( lhs );
results += rhs;
return results;
}
In fact, it's fairly straightforward to provide a template base class
which provides all of these operators automatically. (In this case,
they're declared friend, in order to define them in line in the
template base class.)
Bjarne Stroustrup suggests you use the second form, for cases where you don't modify object itself or when you produce new value/object based on the provided arguments.
I realize this is a basic question but I have searched online, been to cplusplus.com, read through my book, and I can't seem to grasp the concept of overloaded operators. A specific example from cplusplus.com is:
// vectors: overloading operators example
#include <iostream>
using namespace std;
class CVector {
public:
int x,y;
CVector () {};
CVector (int,int);
CVector operator + (CVector);
};
CVector::CVector (int a, int b) {
x = a;
y = b;
}
CVector CVector::operator+ (CVector param) {
CVector temp;
temp.x = x + param.x;
temp.y = y + param.y;
return (temp);
}
int main () {
CVector a (3,1);
CVector b (1,2);
CVector c;
c = a + b;
cout << c.x << "," << c.y;
return 0;
}
From http://www.cplusplus.com/doc/tutorial/classes2/ but reading through it I'm still not understanding them at all. I just need a basic example of the point of the overloaded operator (which I assume is the "CVector CVector::operator+ (CVector param)").
There's also this example from wikipedia:
Time operator+(const Time& lhs, const Time& rhs)
{
Time temp = lhs;
temp.seconds += rhs.seconds;
if (temp.seconds >= 60)
{
temp.seconds -= 60;
temp.minutes++;
}
temp.minutes += rhs.minutes;
if (temp.minutes >= 60)
{
temp.minutes -= 60;
temp.hours++;
}
temp.hours += rhs.hours;
return temp;
}
From "http://en.wikipedia.org/wiki/Operator_overloading"
The current assignment I'm working on I need to overload a ++ and a -- operator.
Thanks in advance for the information and sorry about the somewhat vague question, unfortunately I'm just not sure on it at all.
Operator overloading is the technique that C++ provides to let you define how the operators in the language can be applied to non-built in objects.
In you example for the Time class operator overload for the + operator:
Time operator+(const Time& lhs, const Time& rhs);
With that overload, you can now perform addition operations on Time objects in a 'natural' fashion:
Time t1 = some_time_initializer;
Time t2 = some_other_time_initializer;
Time t3 = t1 + t2; // calls operator+( t1, t2)
The overload for an operator is just a function with the special name "operator" followed by the symbol for the operator being overloaded. Most operators can be overloaded - ones that cannot are:
. .* :: and ?:
You can call the function directly by name, but usually don't (the point of operator overloading is to be able to use the operators normally).
The overloaded function that gets called is determined by normal overload resolution on the arguments to the operator - that's how the compiler knows to call the operator+() that uses the Time argument types from the example above.
One additional thing to be aware of when overloading the ++ and -- increment and decrement operators is that there are two versions of each - the prefix and the postfix forms. The postfix version of these operators takes an extra int parameter (which is passed 0 and has no purpose other than to differentiate between the two types of operator). The C++ standard has the following examples:
class X {
public:
X& operator++(); //prefix ++a
X operator++(int); //postfix a++
};
class Y { };
Y& operator++(Y&); //prefix ++b
Y operator++(Y&, int); //postfix b++
You should also be aware that the overloaded operators do not have to perform operations that are similar to the built in operators - being more or less normal functions they can do whatever you want. For example, the standard library's IO stream interface uses the shift operators for output and input to/from streams - which is really nothing like bit shifting. However, if you try to be too fancy with your operator overloads, you'll cause much confusion for people who try to follow your code (maybe even you when you look at your code later).
Use operator overloading with care.
An operator in C++ is just a function with a special name. So instead of saying Add(int,int) you say operator +(int,int).
Now as any other function, you can overload it to say work on other types. In your vector example, if you overload operator + to take CVector arguments (ie. operator +(CVector, CVector)), you can then say:
CVector a,b,res;
res=a+b;
Since ++ and -- are unary (they take only one argument), to overload them you'd do like:
type operator ++(type p)
{
type res;
res.value++;
return res;
}
Where type is any type that has a field called value. You get the idea.
What you found in those references are not bad examples of when you'd want operator overloading (giving meaning to vector addition, for example), but they're horrible code when it comes down to the details.
For example, this is much more realistic, showing delegating to the compound assignment operator and proper marking of a const member function:
class Vector2
{
double m_x, m_y;
public:
Vector2(double x, double y) : m_x(x), m_y(y) {}
// Vector2(const Vector2& other) = default;
// Vector2& operator=(const Vector2& other) = default;
Vector2& operator+=(const Vector2& addend) { m_x += addend.m_x; m_y += addend.m_y; return *this; }
Vector2 operator+(const Vector2& addend) const { Vector2 sum(*this); return sum += addend; }
};
From your comments above, you dont see the point of all this operator overloading?
Operator overloading is simply 'syntactic sugar' hiding a method call, and making code somehwhat clearer in many cases.
Consider a simple Integer class wrapping an int. You would write add and other arithmetic methods, possibly increment and decrement as well, requiring a method call such as my_int.add(5). now renaming the add method to operator+ allows my_int + 5, which is more intuitive and clearer, cleaner code. But all it is really doing is hiding a call to your operator+ (renamed add?) method.
Things do get a bit more complex though, as operator + for numbers is well understood by everyone above 2nd grade. But as in the string example above, operators should usually only be applied where they have an intuitive meaning. The Apples example is a good example of where NOT to overload operators.
But applied to say, a List class, something like myList + anObject, should be intuitively understood as 'add anObject to myList', hence the use of the + operator. And operator '-' as meaning 'Removal from the list'.
As I said above, the point of all this is to make code (hopefully) clearer, as in the List example, which would you rather code? (and which do you find easier to read?) myList.add( anObject ) or myList + onObject? But in the background, a method (your implementation of operator+, or add) is being called either way. You can almost think of the compiler rewritting the code: my_int + 5 would become my_int.operator+(5)
All the examples given, such as Time and Vector classes, all have intuitive definitions for the operators. Vector addition... again, easier to code (and read) v1 = v2 + v3 than v1 = v2.add(v3). This is where all the caution you are likely to read regarding not going overboard with operators in your classes, because for most they just wont make sense. But of course there is nothing stopping you putting an operator & into a class like Apple, just dont expect others to know what it does without seeing the code for it!
'Overloading' the operator simply means your are supplying the compiler with another definition for that operator, applied to instances of your class. Rather like overloading methods, same name... different parameters...
Hope this helps...
The "operator" in this case is the + symbol.
The idea here is that an operator does something. An overloaded operator does something different.
So, in this case, the '+' operator, normally used to add two numbers, is being "overloaded" to allow for adding vectors or time.
EDIT: Adding two integers is built-in to c++; the compiler automatically understands what you mean when you do
int x, y = 2, z = 2;
x = y + z;
Objects, on the other hand, can be anything, so using a '+' between two objects doesn't inherently make any sense. If you have something like
Apple apple1, apple2, apple3;
apple3 = apple1 + apple2;
What does it mean when you add two Apple objects together? Nothing, until you overload the '+' operator and tell the compiler what it is that you mean when you add two Apple objects together.
An overloaded operator is when you use an operator to work with types that C++ doesn't "natively" support for that operator.
For example, you can typically use the binary "+" operator to add numeric values (floats, ints, doubles, etc.). You can also add an integer type to a pointer - for instance:
char foo[] = "A few words";
char *p = &(foo[3]); // Points to "e"
char *q = foo + 3; // Also points to "e"
But that's it! You can't do any more natively with a binary "+" operator.
However, operator overloading lets you do things the designers of C++ didn't build into the language - like use the + operator to concatenate strings - for instance:
std::string a("A short"), b(" string.");
std::string c = a + b; // c is "A short string."
Once you wrap your head around that, the Wikipedia examples will make more sense.
A operator would be "+", "-" or "+=". These perform different methods on existing objects. This in fact comes down to a method call. Other than normal method calls these look much more natural to a human user. Writing "1 + 2" just looks more normal and is shorter than "add(1,2)". If you overload an operator, you change the method it executes.
In your first example, the "+" operator's method is overloaded, so that you can use it for vector-addition.
I would suggest that you copy the first example into an editor and play a little around with it. Once you understand what the code does, my suggestion would be to implement vector subtraction and multiplication.
Before starting out, there are many operators out there! Here is a list of all C++ operators: list.
With this being said, operator overloading in C++ is a way to make a certain operator behave in a particular way for an object.
For example, if you use the increment/decrement operators (++ and --) on an object, the compiler will not understand what needs to be incremented/decremented in the object because it is not a primitive type (int, char, float...). You must define the appropriate behavior for the compiler to understand what you mean. Operator overloading basically tells the compiler what must be accomplished when the increment/decrement operators are used with the object.
Also, you must pay attention to the fact that there is postfix incrementing/decrementing and prefix incrementing/decrementing which becomes very important with the notion of iterators and you should note that the syntax for overloading these two type of operators is different from each other. Here is how you can overload these operators: Overloading the increment and decrement operators
The accepted answer by Michael Burr is quite good in explaining the technique, but from the comments it seems that besides the 'how' you are interested in the 'why'. The main reasons to provide operator overloads for a given type are improving readability and providing a required interface.
If you have a type for which there is a single commonly understood meaning for an operator in the domain of your problem, then providing that as an operator overload makes code more readable:
std::complex<double> a(1,2), b(3,4), c( 5, 6 );
std::complex<double> d = a + b + c; // compare to d = a.add(b).add(c);
std::complex<double> e = (a + d) + (b + c); // e = a.add(d).add( b.add(c) );
If your type has a given property that will naturally be expressed with an operator, you can overload that particular operator for your type. Consider for example, that you want to compare your objects for equality. Providing operator== (and operator!=) can give you a simple readable way of doing so. This has the advantage of fulfilling a common interface that can be used with algorithms that depend on equality:
struct type {
type( int x ) : value(x) {}
int value;
};
bool operator==( type const & lhs, type const & rhs )
{ return lhs.value == rhs.value; }
bool operator!=( type const & lhs, type const & rhs )
{ return !lhs == rhs; }
std::vector<type> getObjects(); // creates and fills a vector
int main() {
std::vector<type> objects = getObjects();
type t( 5 );
std::find( objects.begin(), objects.end(), t );
}
Note that when the find algorithm is implemented, it depends on == being defined. The implementation of find will work with primitive types as well as with any user defined type that has an equality operator defined. There is a common single interface that makes sense. Compare that with the Java version, where comparison of object types must be performed through the .equals member function, while comparing primitive types can be done with ==. By allowing you to overload the operators you can work with user defined types in the same way that you can with primitive types.
The same goes for ordering. If there is a well defined (partial) order in the domain of your class, then providing operator< is a simple way of implementing that order. Code will be readable, and your type will be usable in all situations where a partial order is required, as inside associative containers:
bool operator<( type const & lhs, type const & rhs )
{
return lhs < rhs;
}
std::map<type, int> m; // m will use the natural `operator<` order
A common pitfall when operator overloading was introduced into the language is that of the 'golden hammer' Once you have a golden hammer everything looks like a nail, and operator overloading has been abused.
It is important to note that the reason for overloading in the first place is improving readability. Readability is only improved if when a programmer looks at the code, the intentions of each operation are clear at first glance, without having to read the definitions. When you see that two complex numbers are being added like a + b you know what the code is doing. If the definition of the operator is not natural (you decide to implement it as adding only the real part of it) then code will become harder to read than if you had provided a (member) function. If the meaning of the operation is not well defined for your type the same happens:
MyVector a, b;
MyVector c = a + b;
What is c? Is it a vector where each element i is the sum of of the respective elements from a and b, or is it a vector created by concatenating the elements of a before the elements of b. To understand the code, you would need to go to the definition of the operation, and that means that overloading the operator is less readable than providing a function:
MyVector c = append( a, b );
The set of operators that can be overloaded is not restricted to the arithmetic and relational operators. You can overload operator[] to index into a type, or operator() to create a callable object that can be used as a function (these are called functors) or that will simplify usage of the class:
class vector {
public:
int operator[]( int );
};
vector v;
std::cout << v[0] << std::endl;
class matrix {
public:
int operator()( int row, int column );
// operator[] cannot be overloaded with more than 1 argument
};
matrix m;
std::cout << m( 3,4 ) << std::endl;
There are other uses of operator overloading. In particular operator, can be overloaded in really fancy ways for metaprogramming purposes, but that is probably much more complex than what you really care for now.
Another use of operator overloading, AFAIK unique to C++, is the ability to overload the assignment operator. If you have:
class CVector
{
// ...
private:
size_t capacity;
size_t length;
double* data;
};
void func()
{
CVector a, b;
// ...
a = b;
}
Then a.data and b.data will point to the same location, and if you modify a, you affect b as well. That's probably not what you want. But you can write:
CVector& CVector::operator=(const CVector& rhs)
{
delete[] data;
capacity = length = rhs.length;
data = new double[length];
memcpy(data, rhs.data, length * sizeof(double));
return (*this);
}
and get a deep copy.
Operator overloading allows you to give own meaning to the operator.
For example, consider the following code snippet:
char* str1 = "String1";
char* str2 = "String2";
char str3[20];
str3 = str1 + str2;
You can overload the "+" operator to concatenate two strings. Doesn't this look more programmer-friendly?
Is there a canonical or recommended pattern for implementing arithmetic operator overloading in C++ number-like classes?
From the C++ FAQ, we have an exception-safe assignment operator that avoids most problems:
class NumberImpl;
class Number {
NumberImpl *Impl;
...
};
Number& Number::operator=(const Number &rhs)
{
NumberImpl* tmp = new NumberImpl(*rhs.Impl);
delete Impl;
Impl = tmp;
return *this;
}
But for other operators (+, +=, etc..) very little advice is given other than to make them behave like the operators on built-in types.
Is there a standard way of defining these? This is what I've come up with - are there pitfalls I'm not seeing?
// Member operator
Number& Number::operator+= (const Number &rhs)
{
Impl->Value += rhs.Impl->Value; // Obviously this is more complicated
return *this;
}
// Non-member non-friend addition operator
Number operator+(Number lhs, const Number &rhs)
{
return lhs += rhs;
}
In Bjarne Stroustrup's book "The C++ Programming Language", in chapter 11 (the one devoted to Operator Overloading) he goes through witting a class for a complex number type (section 11.3).
One thing I do notice from that section is that he implements mixed type operations... this is probably expected for any numeric class.
In general, what you've got looks good.
The big thing to consider when writing any operator is that member operators do not undergo conversions on the left parameter:
struct example {
example(int);
example operator + (example);
};
void foo() {
example e(3), f(6);
e + 4; // okay: right operand is implicitly converted to example
e + f; // okay: no conversions needed.
6 + e; // BAD: no matching call.
}
This is because conversion never applies to this for member functions, and this extends to operators. If the operator was instead example operator + (example, example) in the global namespace, it would compile (or if pass-by-const-ref was used).
As a result, symmetric operators like + and - are generally implemented as non-members, whereas the compound assignment operators like += and -= are implemented as members (they also change data, meaning they should be members). And, since you want to avoid code duplication, the symmetric operators can be implemented in terms of the compound assignment ones (as in your code example, although convention recommends making the temporary inside the function).
The convention is to write operator+(const T&) and operator-(const T&) in terms of operator+=(const T&) and operator-=(const T&). If it makes sense to add and subtract to/from primitive types then you should write a constructor that constructs the object from the primitive type. Then the overloaded operators will also work for primitive types, because the compiler will call the appropriate constructor implicitly.
As you mentioned yourself, you should avoid giving access privileges to functions that don't need it. But in your code above for operator+(Number, const Number&) I'd personally make both parameters const references and use a temp. I think it isn't surprising the commenter below your question missed this; unless you have a good reason not to, avoid surprises and tricks and be as obvious as possible.
If you want your code to integrate with other numeric types, say std::complex, watch out for cyclic conversions. That is, don't supply operator OtherNumeric() in Numeric if OtherNumeric supplies a constructor that takes a Numeric parameter.
It is traditional to write the operator X in terms of the operator =X
It is also traditional the all parameters to the standard operators are const
// Member operator
// This was OK
Number& Number::operator+= (Number const& rhs)
{
Impl->Value += rhs.Impl->Value; // Obviously this is more complicated
return *this;
}
// Non-member non-friend addition operator
Number operator+(Number const& lhs,Number const& rhs)
{
// This I would set the lhs side to const.
// Make a copy into result.
// Then use += add the rhs
Number result(lhs);
return result += rhs;
}
You mention assignment operator.
But you did not mention the copy constructor. Since your class has ownership of a RAW pointer I would expect you to define this as well. The Assignment operator is then traditionally written in terms of the copy constructor.