I am writing some C++ code that needs to test string and character equality, and for the sake of simplicity I'd like to consider the n-dash (0x96) and m-dash (0x97) characters to be identical.
My first instinct was to redefine the equality operator, and started to code, but then ran into a problem:
inline bool operator==(char lhs, char rhs) {
if (lhs == 0x96 && rhs == 0x97) return true; // works fine
else if (lhs == 0x97 && rhs == 0x96) return true; // works fine
else return lhs == rhs; // infinite recursion...
}
In the final line of that function, ideally I'd like to be able to call 'old' form of the equality operator, similar to how a derived class is able to call a base class' version of a function.
I am wondering if this is possible in C++? If not, I'm assuming I should just extract the above code into a separate function and call the function rather than using the operator.
You can't. Once you overload an operator, you replace the default one. (There is an interesting exception: namely std::addressof can be used to circumvent an overloaded & operator).
I'd have strong reservations about overloading operator==(char, char): you'll break a lot of code.
If you really must do it though, you could always write (int)lhs == rhs; which will cause conversion of both operators to int so blocking the recursion. Since int is a superset of char, this will always be defined. Oddly enough, that's why your two prior comparisons work: an implicit conversion of char is taking place which stops the function from calling itself.
First of all, never attempt to overload operators purely on builtin types... Use a function instead...
Your first comparison statements worked because of type promotion
char will be converted to an int and comparisons will be made... the last one calls your operator again
I have a weird error when I try to compile this piece of code.
I will explain my problem.
I defined a vector2D as following:
typedef struct LX_Vector2D
{
float vx;
float vy;
LX_Vector2D& operator =(const LX_Vector2D v); // Defined
} LX_Vector2D;
I also defined two operators on this vector:
bool operator ==(LX_Vector2D& u,LX_Vector2D& v); // Are u and v equal?
LX_Vector2D operator -(LX_Vector2D& u); // Get the opposite vector
All of these overloaded operators was defined.
So I tested these operators in the following code:
LX_Vector2D u = {3.14,-2.56};
LX_Vector2D expected_vec = {-u.vx,-u.vy};
if(expected_vec == (-u)) // ERROR
cout << "OK" << endl;
else
cout << "NO" << endl;
When I compile this code, I have this error:
no match for ‘operator==’ in ‘expected_vec == operator-((* & u))‘
I have no problem with '=' and '==' because I defined and tested them before I implemented '-'.
But when I modify this code to get this:
u = -u;
if(expected_vec == u) // OK
I have no error.
I do not understand that, because it seems these two pieces of code are semantically identical.
Here is the definition of operator '-':
LX_Vector2D operator -(LX_Vector2D& u)
{
return {-u.vx,-u.vy};
}
So my question is:
Why doesn't my compiler recognize ‘expected_vec == (-u)‘ as a call of operator '==' with expected_vec and (-u) as parameters?
Another question:
How can I have the possibility to use if(expected_vec == (-u)) without any problem, if it is possible?
I use g++ 4.6.1.
The problem here is that the result from operator- when used as part of another expression is a temporary value, and that operator== takes non-constant references. A non-constant reference can't bind to a temporary value.
The simple solution? Make the operator== function take constant references:
bool operator ==(const LX_Vector2D& u, const LX_Vector2D& v)
// ^^^^^ ^^^^^
// Note the use of `const`
As a general recommendation, when declaring functions that will not modify their arguments, always pass the arguments as constant. It will avoid problems like this, and may also help the compiler with possible optimizations.
Operator - returns a temporary object:
LX_Vector2D operator -(LX_Vector2D& u)
while your comparison operator accepts a non-const reference:
bool operator ==(LX_Vector2D& u,LX_Vector2D& v)
Temporary objects, like returned by the - operator, cannot be used as non-const references. It's not allowed because there is no purpose modifying an object that is about to go out of scope anyway, so the compiler makes sure you don't even attempt it.
As a general rule, you should make any function that does not modify its arguments take const references instead, especially comparison functions:
bool operator ==(const LX_Vector2D& u,const LX_Vector2D& v)
in addition to the other answers, your assignment operator should also take a const& as in:
LX_Vector2D& operator =(const LX_Vector2D& v)
note the & after the parameter type.
As a general rule, and to avoid the construction of unnecessary copies of objects, parameters of complex types should almost always be const & if you are not planning on changing the parameter instance. Or if you are planning on changing the parameter instance, then simply as a reference, i.e., &.
I'm having some difficulty with this. I've determined I need to overload this operator for my personal project. It is necessitated by the use of the following line:
if(playerVec[i] == 0)
The player class has several data members for calculating one particular data member, mInitiative. This is the one I want to check in my if condition. Here is my attempt at overloading it:
bool operator==(const Player& lhs) const {
return mInitiative == lhs.mInitiative;
}
It seems fine enough, but the error persists. If I want to compare that particular player datum to an integer (in this case, 0), how do I go about it? What's the mistake in my approach?
EDIT: I have tried:
bool operator==(const Player& lhs, int rhs) const {
//...
}
But the compiler says there are too many parameters for the function. Why is this? Shouldn't == be able to take two?
Thanks!
There are two ways to overload an equality operator: declare it as a member, taking one argument (rhs); or declare it as a global, taking two arguments (lhs and rhs). Since your lhs is a Player, and your rhs is an integer, here are the two ways to define it:
// declared inside Player class as a member
bool operator == (int rhs) const
{
return mInitiative == rhs;
}
// can also be declared inside Player class, but is not a member due to friend keyword
friend bool operator == (Player const& lhs, int rhs)
{
return lhs.mInitiative == rhs;
}
That is leaving aside the style considerations of overloading operators in such a way.
When trying to overload equality operator (i.e. ==), you always need to think about whether the target instances are really the same.
In your case, I think people might be confused when reading the following code if you provide the Player to integer comparison. Since it looks like checking whether a pointer is null or not:
if(playerVec[i] == 0)
Rather than overloading == operator of Player to compare with integer, I would suggest providing a get() function, which allows you to compare Player with integer more clearly. For example:
if (playerVec[i].getPlayerID() == 0)
If you will use some stl function to manage your Player vector (eg. sorting), then you can overload == or > operator for two Player instances.
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?