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 have a couple of pure virtual classes, Matrix and Vector. Throughout my code base I try to only create dependencies on them and not their concrete subclasses e.g. SimpleTransformationMatrix44 and SimpleTranslationVector4. The motivation for this is that I can use third party (adapted) classes in place of mine without much trouble.
I would like to overload the arithmetic operators (sourced from here):
T T::operator +(const T& b) const;
T T::operator -(const T& b) const;
T T::operator *(const T& b) const;
I want to declare them in the pure virtual classes so that it is valid to perform the operations on references/pointers to them, the problem being that an abstract class cannot be returned by value. The best solution I can think of is something like this:
std::unique_ptr<T> T::operator +(const T& b) const;
std::unique_ptr<T> T::operator -(const T& b) const;
std::unique_ptr<T> T::operator *(const T& b) const;
Which allows this (without down casts!):
std::unique_ptr<Matrix> exampleFunction(const Matrix& matrix1, const Matrix& matrix2)
{
std::unique_ptr<Matrix> product = matrix1 * matrix2;
return std::move(product);
}
A pointer seems to be the only option in this case since returning a value is invalid and returning a reference is just plain silly.
So I guess my question is: Have I gone off the plot with this idea? If you saw it in some code you were working on would you be thinking WTF to yourself? Is there a better way to achieve this?
First off: Overloading operators is something that applies best to value types. As you have found out, polymorphism doesn't play well with it. If you are willing to walk on crutches, this might help, though:
If you follow the advice of Stackoverflow's operator overloading FAQ, you will implement operator+() as a non-member atop of operator+=(). The latter returns a reference. It's still a problem, because it can only return a base class reference, but as long as you only use it for expressions expecting that, you are fine.
If you then templatize operator+() as DeadMG suggested, you might be able to do what you want:
template<typename T>
T operator+(const T lhs, const T& rhs)
{
lhs += rhs;
return lhs;
}
Note that this would catch any T for which no better-matching operator+() overload can be found. (This might seem like a good idea — until you forget to include a header and this operator catches the x+y, making the code compile, but silently produce wrong results.) So you might want to restrict this.
One way would be to put it into the same namespace as you matrix and vector types. Or you use a static_assert to ensure only types derived from the two are passed in.
Have I gone off the plot with this idea?
Yes. The appropriate solution to this problem is to implement flexibility via template, not inheritance. Inheritance is most definitely not suited to this kind of problem. In addition, it's usual (if not mandated) to have the dimensions of the vector or matrix specified as a template parameter in addition, instead of at run-time.
I'm compiling some c++ code of a class MegaInt which is a positive decimal type class that allows arithmetic operations on huge numbers.
I want to overload operator bool to allow code like this:
MegaInt m(45646578676547676);
if(m)
cout << "YaY!" << endl;
This is what I did:
header:
class MegaInt
{
public:
...
operator bool() const;
};
const MegaInt operator+(const MegaInt & left, const MegaInt & right);
const MegaInt operator*(const MegaInt & left, const MegaInt & right);
implementation:
MegaInt::operator bool() const
{
return *this != 0;
}
const MegaInt operator+(const MegaInt & left, const MegaInt & right)
{
MegaInt ret = left;
ret += right;
return ret;
}
Now, the problem is if I do:
MegaInt(3424324234234342) + 5;
It gives me this error:
ambiguous overload for 'operator+' in 'operator+(const MegaInt&, const MegaInt&)
note: candidates are: operator+(int, int) |
note: const MegaInt operator+(const MegaInt&, const MegaInt&)|
I don't know why. How is the overloaded bool() causing operator+ to become ambiguous?¸
Thank You.
Well, everyone gave me great answers, unfortunately, none of them seem to solve my problem entirely.
Both void* or the Safe Bool Idiom works. Except for one tiny problem, I hope has a workaround:
When comparing with 0 like:
if (aMegaInt == 0)
The compiler gives an ambiguous overload error again. I understand why: it doesn't know if we're comparing to false or to MegaInt of value 0. None the less, in that case, I'd want it to cast to MegaInt(0). Is there a way to force this?
Thank You Again.
The C++ compiler is allowed to automatically convert bool into int for you, and that's what it wants to do here.
The way to solve this problem is to employ the safe bool idiom.
Technically, creating an operator void* is not an example of the safe bool idiom, but it's safe enough in practice, because the bool/int problem you're running into is a common error, and messes up some perfectly reasonable and otherwise correct code (as you see from your question), but misuses of the void* conversion are not so common.
The wikipedia entry on explicit conversion operators for C++0x has a decent summary of why you see this error pre-C++0x. Basically, the bool conversion operator is an integral conversion type, so it will be used in an integral arithmetic expression. The pre-C++0x fix is to instead use void * as the conversion operator; void * can be converted to a boolean expression, but not to an integral expression.
As Erik's answer states, the problem here is that by providing an implicit conversion to bool you are opening the door to expressions that can mean multiple things; in this case the compiler will complain of ambiguity and give your an error.
However, note that providing an implicit conversion to void* will not let you off the hook; it will just change the set of expressions which present a problem.
There are two airtight solutions to this issue:
Make the conversion to bool explicit (which can be undesirable if the class represents an entity with an intuitive "true/false" value)
Use the safe bool idiom (this really covers all bases, but as many good things in life and C++ is way too complicated -- you pay the price)
The problem is that bool can freely convert to int. So the expression MegaInt(3424324234234342) + 5; can equally validly be interpreted this way:
(bool)(MegaInt(3424324234234342)) + 5;
or:
MegaInt(3424324234234342) + MegaInt(5);
Each one of those expressions involves one user defined conversion and are equal in the eyes of the compiler. Conversion to bool is highly problematic for this reason. It would be really nice to have a way to say it should only happen in a context that explicitly requires a bool, but there isn't. :-/
The conversion to void * that someone else suggests is a workaround, but I think as a workaround it has problems of its own and I wouldn't do it.
MegaInt(3424324234234342) + 5;
MegaInt + int;
Should the compiler convert your MegaInt to an integral (bool is an integral type) or the integer to MegaInt (you have an int constructor)?
You fix this by creating an operator void * instead of an operator bool:
operator void *() const { return (*this != 0) ? ((void *) 1) : ((void *) 0); }
Others have mentioned the Safe Bool Idiom. However, for objects like yours it is a bad idea to add all this nasty, special logic when you want full algebra support anyway.
You're defining a custom integer type. You get far more for your effort by defining "operator==" and "operator!=", then implementing "operator bool()" as something like:
operator bool()
{
return (*this != 0);
}
Just from those 3 functions you get all of the "if" idioms for integers, and they'll behave the same for your custom ints as the built-in ones: "if(a==b)", "if(a!=b)", "if(a)", "if(!a)". Your implicit "bool" rule will also (if you're careful) work intuitively as well.
Besides, the full "Safe Bool Idiom" is unnecessary. Think about it- the only time you need it is "1) comparison of 2 objects is ill-defined or undefined, 2) cast to (int) or other primitive types needs to be protected and 3) object validity IS well-defined (the actual source of the returned bool)."
Well, 2) is only a consideration if you actually wish to SUPPORT casting to a numeric type like int or float. But for objects that have NO well-defined notion of equality (# 1), providing such casts unavoidably creates the risk of the very "if(a==b)" logic bombs the idiom supposedly protects you from. Just declare "operator int()" and such private like you do with the copy ctor on non-copyable objects and be done with it:
class MyClass {
private:
MyClass(const MyClass&);
operator int();
operator long();
// float(), double(), etc. ...
public:
// ctor & dtor ..
bool operator==(const MyClass& other) const { //check for equality logic... }
bool operator!=(const MyClass& other) const { return !(*this == other); }
operator bool() { return (*this != 0); }
};
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?