What are the basic rules and idioms for operator overloading?

Asked 2023-09-20 20:30:22 View 277,507

Note: The answers were given in a specific order, but since many users sort answers according to votes, rather than the time they were given, here's an index of the answers in the order in which they make the most sense:

(Note: This is meant to be an entry to Stack Overflow's C++ FAQ. If you want to critique the idea of providing an FAQ in this form, then the posting on meta that started all this would be the place to do that. Answers to that question are monitored in the C++ chatroom, where the FAQ idea started in the first place, so your answer is very likely to get read by those who came up with the idea.)

Answers

Common operators to overload

Most of the work in overloading operators is boiler-plate code. That is little wonder, since operators are merely syntactic sugar, their actual work could be done by (and often is forwarded to) plain functions. But it is important that you get this boiler-plate code right. If you fail, either your operator’s code won’t compile or your users’ code won’t compile or your users’ code will behave surprisingly.

Assignment Operator

There's a lot to be said about assignment. However, most of it has already been said in GMan's famous Copy-And-Swap FAQ, so I'll skip most of it here, only listing the perfect assignment operator for reference:

X& X::operator=(X rhs)
{
  swap(rhs);
  return *this;
}

Bitshift Operators (used for Stream I/O)

The bitshift operators << and >>, although still used in hardware interfacing for the bit-manipulation functions they inherit from C, have become more prevalent as overloaded stream input and output operators in most applications. For guidance overloading as bit-manipulation operators, see the section below on Binary Arithmetic Operators. For implementing your own custom format and parsing logic when your object is used with iostreams, continue.

The stream operators, among the most commonly overloaded operators, are binary infix operators for which the syntax specifies no restriction on whether they should be members or non-members. Since they change their left argument (they alter the stream’s state), they should, according to the rules of thumb, be implemented as members of their left operand’s type. However, their left operands are streams from the standard library, and while most of the stream output and input operators defined by the standard library are indeed defined as members of the stream classes, when you implement output and input operations for your own types, you cannot change the standard library’s stream types. That’s why you need to implement these operators for your own types as non-member functions. The canonical forms of the two are these:

std::ostream& operator<<(std::ostream& os, const T& obj)
{
  // write obj to stream

  return os;
}

std::istream& operator>>(std::istream& is, T& obj)
{
  // read obj from stream

  if( /* no valid object of T found in stream */ )
    is.setstate(std::ios::failbit);

  return is;
}

When implementing operator>>, manually setting the stream’s state is only necessary when the reading itself succeeded, but the result is not what would be expected.

Function call operator

The function call operator, used to create function objects, also known as functors, must be defined as a member function, so it always has the implicit this argument of member functions. Other than this, it can be overloaded to take any number of additional arguments, including zero.

Here's an example of the syntax:

struct X {
    // Overloaded call operator
    int operator()(const std::string& y) {
        return /* ... */;
    }
};

Usage:

X f;
int a = f("hello");

Throughout the C++ standard library, function objects are always copied. Your own function objects should therefore be cheap to copy. If a function object absolutely needs to use data which is expensive to copy, it is better to store that data elsewhere and have the function object refer to it.

Comparison operators

This section has been moved elsewhere
See this FAQ answer for overloading the binary infix ==, !=, <, >, <=, and >= operators, as well as the <=> three-way comparison, aka. "spaceship operator" in C++20. There is so much to say about comparison operators that it would exceed the scope of this answer.

In the most simple case, you can overload all comparison comparison operators by defaulting <=> in C++20:

#include <compare>

struct X {
  // defines ==, !=, <, >, <=, >=, <=>
  friend auto operator<=>(const X&, const X&) = default;
};

If you can't do this, continue to the linked answer.

Logical operators

The unary prefix negation ! should be implemented as a member function. It is usually not a good idea to overload it because of how rare and surprising it is.

struct X {
  X operator!() const { return /* ... */; }
};

The remaining binary logical operators (||, &&) should be implemented as free functions. However, it is very unlikely that you would find a reasonable use case for these1.

X operator&&(const X& lhs, const X& rhs) { return /* ... */; }
X operator||(const X& lhs, const X& rhs) { return /* ... */; }

1 It should be noted that the built-in version of || and && use shortcut semantics. While the user defined ones (because they are syntactic sugar for method calls) do not use shortcut semantics. User will expect these operators to have shortcut semantics, and their code may depend on it, Therefore it is highly advised NEVER to define them.

Arithmetic Operators

Unary arithmetic operators

The unary increment and decrement operators come in both prefix and postfix flavor. To tell one from the other, the postfix variants take an additional dummy int argument. If you overload increment or decrement, be sure to always implement both prefix and postfix versions. Here is the canonical implementation of increment, decrement follows the same rules:

struct X {
  X& operator++()
  {
    // do actual increment
    return *this;
  }
  X operator++(int)
  {
    X tmp(*this);
    operator++();
    return tmp;
  }
};

Note that the postfix variant is implemented in terms of prefix. Also note that postfix does an extra copy.2

Overloading unary minus and plus is not very common and probably best avoided. If needed, they should probably be overloaded as member functions.

2 Also note that the postfix variant does more work and is therefore less efficient to use than the prefix variant. This is a good reason to generally prefer prefix increment over postfix increment. While compilers can usually optimize away the additional work of postfix increment for built-in types, they might not be able to do the same for user-defined types (which could be something as innocently looking as a list iterator). Once you got used to do i++, it becomes very hard to remember to do ++i instead when i is not of a built-in type (plus you'd have to change code when changing a type), so it is better to make a habit of always using prefix increment, unless postfix is explicitly needed.

Binary arithmetic operators

For the binary arithmetic operators, do not forget to obey the third basic rule operator overloading: If you provide +, also provide +=, if you provide -, do not omit -=, etc. Andrew Koenig is said to have been the first to observe that the compound assignment operators can be used as a base for their non-compound counterparts. That is, operator + is implemented in terms of +=, - is implemented in terms of -= etc.

According to our rules of thumb, + and its companions should be non-members, while their compound assignment counterparts (+= etc.), changing their left argument, should be a member. Here is the exemplary code for += and +; the other binary arithmetic operators should be implemented in the same way:

struct X {
  X& operator+=(const X& rhs)
  {
    // actual addition of rhs to *this
    return *this;
  }
};

inline X operator+(const X& lhs, const X& rhs)
{
  X result = lhs;
  result += lhs;
  return result;
}

operator+= returns its result per reference, while operator+ returns a copy of its result. Of course, returning a reference is usually more efficient than returning a copy, but in the case of operator+, there is no way around the copying. When you write a + b, you expect the result to be a new value, which is why operator+ has to return a new value.3

Also note that operator+ can be slightly shortened by passing lhs by value, not by reference. However, this would be leaking implementation details, make the function signature asymmetric, and would prevent named return value optimization where result is the same object as the one being returned.

Sometimes, it's impractical to implement @ in terms of @=, such as for matrix multiplication. In that case, you can also delegate @= to @:

struct Matrix {
  // you can also define non-member functions inside the class, i.e. "hidden friends"
  friend Matrix operator*(const Matrix& lhs, const Matrix& rhs) {
    Matrix result;
    // do matrix multiplication
    return result;
  }
  Matrix& operator*=(const Matrix& rhs)
  {
    return *this = *this * rhs; // assuming operator= returns a reference
  }
};

The bit manipulation operators ~ & | ^ << >> should be implemented in the same way as the arithmetic operators. However, (except for overloading << and >> for output and input) there are very few reasonable use cases for overloading these.

3 Again, the lesson to be taken from this is that a += b is, in general, more efficient than a + b and should be preferred if possible.

Array Subscripting

The array subscript operator is a binary operator which must be implemented as a class member. It is used for container-like types that allow access to their data elements by a key. The canonical form of providing these is this:

struct X {
        value_type& operator[](index_type idx);
  const value_type& operator[](index_type idx) const;
  // ...
};

Unless you do not want users of your class to be able to change data elements returned by operator[] (in which case you can omit the non-const variant), you should always provide both variants of the operator.

Operators for Pointer-like Types

For defining your own iterators or smart pointers, you have to overload the unary prefix dereference operator * and the binary infix pointer member access operator ->:

struct my_ptr {
        value_type& operator*();
  const value_type& operator*() const;
        value_type* operator->();
  const value_type* operator->() const;
};

Note that these, too, will almost always need both a const and a non-const version. For the -> operator, if value_type is of class (or struct or union) type, another operator->() is called recursively, until an operator->() returns a value of non-class type.

The unary address-of operator should never be overloaded.

For operator->*() see this question. It's rarely used and thus rarely ever overloaded. In fact, even iterators do not overload it.


Continue to Conversion Operators

Answered   2023-09-20 20:30:22

  • operator->() is actually extremely weird. It's not required to return a value_type* -- in fact, it can return another class type, provided that class type has an operator->(), which will then be called subsequently. This recursive calling of operator->()s proceeds until a value_type* return type occurs. Madness! :) - anyone
  • Note that there are cases when defining operatorX in terms of operatorX= if not practical. For matrices and polynomials (english?), code factorization of the multiplication goes the other way: operator*= should be defined in terms of operator*. - anyone
  • It's not exactly about effectiveness. It's about we cannot do it in the traditional-idiomatic way in a (very) few cases: when the definition of both operands need to stay unchanged while we compute the result. And as I said, there are two classical examples: matrices multiplication, and multiplication of polynomials. We could define * in terms of *= but it would be awkward because one of the first operation of *= would to create a new object, result of the computation. Then, after the for-ijk loop, we would swap this temporary object with *this. ie. 1.copy, 2.operator*, 3.swap - anyone
  • I disagree with the const/non-const versions of your pointer-like operators, e.g. ` const value_type& operator*() const;` - this would be like having a T* const returning a const T& on dereferencing, which is not the case. Or in other words: a const pointer does not imply a const pointee. In fact, it is not trivial to mimic T const * - which is the reason for the whole const_iterator stuff in the standard library. Conclusion: the signature should be reference_type operator*() const; pointer_type operator->() const - anyone
  • One comment: The implementation of binary arithmetic operators suggested is not such efficient as it can be. Se Boost operators headers simmetry note: boost.org/doc/libs/1_54_0/libs/utility/operators.htm#symmetry One more copy can be avoided if you use a local copy of the first parameter, do +=, and return the local copy. This enables NRVO optimization. - anyone

The Three Basic Rules of Operator Overloading in C++

When it comes to operator overloading in C++, there are three basic rules you should follow. As with all such rules, there are indeed exceptions. Sometimes people have deviated from them and the outcome was not bad code, but such positive deviations are few and far between. At the very least, 99 out of 100 such deviations I have seen were unjustified. However, it might just as well have been 999 out of 1000. So you’d better stick to the following rules.

  1. 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.
    Basically, the first and foremost rule for overloading operators, at its very heart, says: Don’t do it. That might seem strange, because there is a lot to be known about operator overloading and so a lot of articles, book chapters, and other texts deal with all this. But despite this seemingly obvious evidence, there are only a surprisingly few cases where operator overloading is appropriate. The reason is that actually it is hard to understand the semantics behind the application of an operator unless the use of the operator in the application domain is well known and undisputed. Contrary to popular belief, this is hardly ever the case.

  2. Always stick to the operator’s well-known semantics.
    C++ poses no limitations on the semantics of overloaded operators. Your compiler will happily accept code that implements the binary + operator to subtract from its right operand. However, the users of such an operator would never suspect the expression a + b to subtract a from b. Of course, this supposes that the semantics of the operator in the application domain is undisputed.

  3. Always provide all out of a set of related operations.
    Operators are related to each other and to other operations. If your type supports a + b, users will expect to be able to call a += b, too. If it supports prefix increment ++a, they will expect a++ to work as well. If they can check whether a < b, they will most certainly expect to also to be able to check whether a > b. If they can copy-construct your type, they expect assignment to work as well.


Continue to The Decision between Member and Non-member.

Answered   2023-09-20 20:30:22

  • The only thing of which I am aware which violates any of these is boost::spirit lol. - anyone
  • @Billy: According to some, abusing + for string concatenation is a violation, but it has by now become well established praxis, so that it seems natural. Although I do remember a home-brew string class I saw in the 90ies that used binary & for this purpose (referring to BASIC for established praxis). But, yeah, putting it into the std lib basically set this in stone. The same goes for abusing << and >> for IO, BTW. Why would left-shifting be the obvious output operation? Because we all learned about it when we saw our first "Hello, world!" application. And for no other reason. - anyone
  • @curiousguy: If you have to explain it, it's not obviously clear and undisputed. Likewise if you need to discuss or defend the overloading. - anyone
  • @sbi: "peer review" is always a good idea. To me a badly choosen operator is not different from a badly choosen function name (I saw many). Operator are just functions. No more no less. Rules are just the same. And to understand if an idea is good, the best way is understand how long does it takes to be understood. (Hence, peer review is a must, but peers must be chosen between people free from dogmas and prejudice.) - anyone
  • @sbi To me, the only absolutely obvious and indisputable fact about operator== is that it should be an equivalence relation (IOW, you should not use non signaling NaN). There are many useful equivalence relations on containers. What does equality means? "a equals b" means that a and b have the same mathematical value. The concept of mathematical value of a (non-NaN) float is clear, but the mathematical value of a container can have many distinct (type recursive) useful definitions. The strongest definition of equality is "they are the same objects", and it is useless. - anyone

The Decision between Member and Non-member

Category Operators Decision
Mandatory member functions [], (), =, ->, ... member function (mandated by the C++ standard)
Pointer-to-member access ->* member function
Unary ++, -, *, new, ... member function, except for enumerations
Compound assignment +=, |=, *=, ... member function, except for enumerations
Other operators +, ==, <=>, /, ... prefer non-member

The binary operators = (assignment), [] (array subscription), -> (member access), as well as the n-ary () (function call) operator, must always be implemented as member functions, because the syntax of the language requires them to.

Other operators can be implemented either as members or as non-members. Some of them, however, usually have to be implemented as non-member functions, because their left operand cannot be modified by you. The most prominent of these are the input and output operators << and >>, whose left operands are stream classes from the standard library which you cannot change.

For all operators where you have to choose to either implement them as a member function or a non-member function, use the following rules of thumb to decide:

  1. If it is a unary operator, implement it as a member function.
  2. If a binary operator treats both operands equally (it leaves them unchanged), implement this operator as a non-member function.
  3. If a binary operator does not treat both of its operands equally (usually it will change its left operand), it might be useful to make it a member function of its left operand’s type, if it has to access the operand's private parts.

Of course, as with all rules of thumb, there are exceptions. If you have a type

enum Month {Jan, Feb, ..., Nov, Dec}

and you want to overload the increment and decrement operators for it, you cannot do this as a member functions, since in C++, enum types cannot have member functions. So you have to overload it as a free function. And operator<() for a class template nested within a class template is much easier to write and read when done as a member function inline in the class definition. But these are indeed rare exceptions.

(However, if you make an exception, do not forget the issue of const-ness for the operand that, for member functions, becomes the implicit this argument. If the operator as a non-member function would take its left-most argument as a const reference, the same operator as a member function needs to have a const at the end to make *this a const reference.)


Continue to Common operators to overload.

Answered   2023-09-20 20:30:22

  • Herb Sutter's item in Effective C++ (or is it C++ Coding Standards?) says one should prefer non-member non-friend functions to member functions, to increase the encapsulation of the class. IMHO, the encapsulation reason takes precedence to your rule of thumb, but it does not decrease the quality value of your rule of thumb. - anyone
  • @paercebal: Effective C++ is by Meyers, C++ Coding Standards by Sutter. Which one are you referring to? Anyway, I dislike the idea of, say, operator+=() not being a member. It has to change its left-hand operand, so by definition it has to dig deep into its innards. What would you gain by not making it a member? - anyone
  • @sbi: Item 44 in C++ Coding Standards (Sutter) Prefer writing nonmember nonfriend functions, of course, it only applies if you can actually write this function using only the public interface of the class. If you cannot (or can but it would hinder performance badly), then you have to make it either member or friend. - anyone
  • @sbi : Oops, Effective, Exceptional... No wonder I mix the names up. Anyway the gain is to limit as much as possible the number of functions that have access to an object private/protected data. This way, you increase the encapsulation of your class, making its maintenance/testing/evolution easier. - anyone
  • @sbi : One example. Let's say you're coding a String class, with both the operator += and the append methods. The append method is more complete, because you can append a substring of the parameter from index i to index n -1: append(string, start, end) It seems logical to have += call append with start = 0 and end = string.size. At that moment, append could be a member method, but operator += doesn't need to be a member, and making it a non-member would decrease the quantity of code playing with the String innards, so it is a good thing.... ^_^ ... - anyone

The General Syntax of operator overloading in C++

You cannot change the meaning of operators for built-in types in C++, operators can only be overloaded for user-defined types1. That is, at least one of the operands has to be of a user-defined type. As with other overloaded functions, operators can be overloaded for a certain set of parameters only once.

Not all operators can be overloaded in C++. Among the operators that cannot be overloaded are: . :: sizeof typeid .* and the only ternary operator in C++, ?:

Among the operators that can be overloaded in C++ are these:

Category Operators Arity and Placement
Arithmetic + - * / % and += -= *= /= %= binary infix
+ - unary prefix
++ -- unary prefix and postfix
Bitwise & | ^ << >> and &= |= ^= <<= >>= binary infix
~ unary prefix
Comparison == != < > <= >= <=> binary infix
Logical || && binary infix
! unary prefix
Allocation functions new new[] delete delete[] unary prefix
User-defined conversions T unary
Assignment = binary infix
Member access -> ->* binary infix
Indirection/Address-of * & unary prefix
Function call () N-ary postfix
Subscript [] N-ary2 postfix
Coroutine await co_await unary prefix
Comma , binary infix

However, the fact that you can overload all of these does not mean you should do so. See the basic rules of operator overloading.

In C++, operators are overloaded in the form of functions with special names. As with other functions, overloaded operators can generally be implemented either as a member function of their left operand's type or as non-member functions. Whether you are free to choose or bound to use either one depends on several criteria.3 A unary operator @4, applied to an object x, is invoked either as operator@(x) or as x.operator@(). A binary infix operator @, applied to the objects x and y, is called either as operator@(x,y) or as x.operator@(y).5

Operators that are implemented as non-member functions are sometimes friend of their operand’s type.

1 The term “user-defined” might be slightly misleading. C++ makes the distinction between built-in types and user-defined types. To the former belong for example int, char, and double; to the latter belong all struct, class, union, and enum types, including those from the standard library, even though they are not, as such, defined by users.

2 The subscript operator used to be binary, not N-ary until C++23.

3 This is covered in a later part of this FAQ.

4 The @ is not a valid operator in C++ which is why I use it as a placeholder.

5 The only ternary operator in C++ cannot be overloaded and the only n-ary operator must always be implemented as a member function.


Continue to The Three Basic Rules of Operator Overloading in C++.

Answered   2023-09-20 20:30:22

  • Addendum: Some compilers provide a (typically hidden, internal) switch that allows you to overload operators for non-user-defined types. Any code that redefines built-in operators in this manner is non-compliant. - anyone

Conversion Operators (also known as User Defined Conversions)

In C++ you can create conversion operators, operators that allow the compiler to convert between your types and other defined types. There are two types of conversion operators, implicit and explicit ones.

Implicit Conversion Operators (C++98/C++03 and C++11)

An implicit conversion operator allows the compiler to implicitly convert (like the conversion between int and long) the value of a user-defined type to some other type.

The following is a simple class with an implicit conversion operator:

class my_string {
public:
  operator const char*() const {return data_;} // This is the conversion operator
private:
  const char* data_;
};

Implicit conversion operators, like one-argument constructors, are user-defined conversions. Compilers will grant one user-defined conversion when trying to match a call to an overloaded function.

void f(const char*);

my_string str;
f(str); // same as f( str.operator const char*() )

At first this seems very helpful, but the problem with this is that the implicit conversion even kicks in when it isn’t expected to. In the following code, void f(const char*) will be called because my_string() is not an lvalue, so the first does not match:

void f(my_string&);
void f(const char*);

f(my_string());

Beginners easily get this wrong and even experienced C++ programmers are sometimes surprised because the compiler picks an overload they didn’t suspect. These problems can be mitigated by explicit conversion operators.

Explicit Conversion Operators (C++11)

Unlike implicit conversion operators, explicit conversion operators will never kick in when you don't expect them to. The following is a simple class with an explicit conversion operator:

class my_string {
public:
  explicit operator const char*() const {return data_;}
private:
  const char* data_;
};

Notice the explicit. Now when you try to execute the unexpected code from the implicit conversion operators, you get a compiler error:

prog.cpp: In function ‘int main()’:
prog.cpp:15:18: error: no matching function for call to ‘f(my_string)’
prog.cpp:15:18: note: candidates are:
prog.cpp:11:10: note: void f(my_string&)
prog.cpp:11:10: note:   no known conversion for argument 1 from ‘my_string’ to ‘my_string&’
prog.cpp:12:10: note: void f(const char*)
prog.cpp:12:10: note:   no known conversion for argument 1 from ‘my_string’ to ‘const char*’

To invoke the explicit cast operator, you have to use static_cast, a C-style cast, or a constructor style cast ( i.e. T(value) ).

However, there is one exception to this: The compiler is allowed to implicitly convert to bool. In addition, the compiler is not allowed to do another implicit conversion after it converts to bool (a compiler is allowed to do 2 implicit conversions at a time, but only 1 user-defined conversion at max).

Because the compiler will not cast "past" bool, explicit conversion operators now remove the need for the Safe Bool idiom. For example, smart pointers before C++11 used the Safe Bool idiom to prevent conversions to integral types. In C++11, the smart pointers use an explicit operator instead because the compiler is not allowed to implicitly convert to an integral type after it explicitly converted a type to bool.

Continue to Overloading new and delete.

Answered   2023-09-20 20:30:22

Overloading new and delete operators

Note: This only deals with the syntax of overloading new and delete, not with the implementation of such overloaded operators. I think that the semantics of overloading new and delete deserve their own FAQ, within the topic of operator overloading I can never do it justice.

Basics

In C++, when you write a new expression like new T(arg) two things happen when this expression is evaluated: First operator new is invoked to obtain raw memory, and then the appropriate constructor of T is invoked to turn this raw memory into a valid object. Likewise, when you delete an object, first its destructor is called, and then the memory is returned to operator delete.
C++ allows you to tune both of these operations: memory management and the construction/destruction of the object at the allocated memory. The latter is done by writing constructors and destructors for a class. Fine-tuning memory management is done by writing your own operator new and operator delete.

The first of the basic rules of operator overloading – don’t do it – applies especially to overloading new and delete. Almost the only reasons to overload these operators are performance problems and memory constraints, and in many cases, other actions, like changes to the algorithms used, will provide a much higher cost/gain ratio than attempting to tweak memory management.

The C++ standard library comes with a set of predefined new and delete operators. The most important ones are these:

void* operator new(std::size_t) throw(std::bad_alloc); 
void  operator delete(void*) throw(); 
void* operator new[](std::size_t) throw(std::bad_alloc); 
void  operator delete[](void*) throw(); 

The first two allocate/deallocate memory for an object, the latter two for an array of objects. If you provide your own versions of these, they will not overload, but replace the ones from the standard library.
If you overload operator new, you should always also overload the matching operator delete, even if you never intend to call it. The reason is that, if a constructor throws during the evaluation of a new expression, the run-time system will return the memory to the operator delete matching the operator new that was called to allocate the memory to create the object in. If you do not provide a matching operator delete, the default one is called, which is almost always wrong.
If you overload new and delete, you should consider overloading the array variants, too.

Placement new

C++ allows new and delete operators to take additional arguments.
So-called placement new allows you to create an object at a certain address which is passed to:

class X { /* ... */ };
char buffer[ sizeof(X) ];
void f()
{ 
  X* p = new(buffer) X(/*...*/);
  // ... 
  p->~X(); // call destructor 
} 

The standard library comes with the appropriate overloads of the new and delete operators for this:

void* operator new(std::size_t,void* p) throw(std::bad_alloc); 
void  operator delete(void* p,void*) throw(); 
void* operator new[](std::size_t,void* p) throw(std::bad_alloc); 
void  operator delete[](void* p,void*) throw(); 

Note that, in the example code for placement new given above, operator delete is never called, unless the constructor of X throws an exception.

You can also overload new and delete with other arguments. As with the additional argument for placement new, these arguments are also listed within parentheses after the keyword new. Merely for historical reasons, such variants are often also called placement new, even if their arguments are not for placing an object at a specific address.

Class-specific new and delete

Most commonly you will want to fine-tune memory management because measurement has shown that instances of a specific class, or of a group of related classes, are created and destroyed often and that the default memory management of the run-time system, tuned for general performance, deals inefficiently in this specific case. To improve this, you can overload new and delete for a specific class:

class my_class { 
  public: 
    // ... 
    void* operator new(std::size_t);
    void  operator delete(void*);
    void* operator new[](std::size_t);
    void  operator delete[](void*);
    // ...  
}; 

Overloaded thus, new and delete behave like static member functions. For objects of my_class, the std::size_t argument will always be sizeof(my_class). However, these operators are also called for dynamically allocated objects of derived classes, in which case it might be greater than that.

Global new and delete

To overload the global new and delete, simply replace the pre-defined operators of the standard library with our own. However, this rarely ever needs to be done.

Answered   2023-09-20 20:30:22

  • I also don't agree that replacing the global operator new and delete is usually for performance: on the contrary, it's usually for bug tracing. - anyone
  • You should also note, that if you use an overloaded new operator you're required to also provide a delete operator with matching arguments. You say that in the section on global new/delete where it isn't of much interest. - anyone
  • @Yttrill you are confusing things. The meaning gets overloaded. What "operator overloading" means is that the meaning is overloaded. It does not mean that literally functions are overloaded, and in particular operator new will not overload the Standard's version. @sbi doesn't claim the opposite. It's common to call it "overloading new" much as it is common to say "overloading addition operator". - anyone
  • @sbi: See (or better, link to) gotw.ca/publications/mill15.htm . It is only good practice towards people which sometimes use nothrow new. - anyone
  • "If you do not provide a matching operator delete, the default one is called" -> Actually, if you add any arguments and do not create a matching delete, no operator delete is called at all, and you have a memory leak. (15.2.2, the storage occupied by the object is deallocated only if an appropriate ... operator delete is found) - anyone

Why can't operator<< function for streaming objects to std::cout or to a file be a member function?

Let's say you have:

struct Foo
{
   int a;
   double b;

   std::ostream& operator<<(std::ostream& out) const
   {
      return out << a << " " << b;
   }
};

Given that, you cannot use:

Foo f = {10, 20.0};
std::cout << f;

Since operator<< is overloaded as a member function of Foo, the LHS of the operator must be a Foo object. Which means, you will be required to use:

Foo f = {10, 20.0};
f << std::cout

which is very non-intuitive.

If you define it as a non-member function,

struct Foo
{
   int a;
   double b;
};

std::ostream& operator<<(std::ostream& out, Foo const& f)
{
   return out << f.a << " " << f.b;
}

You will be able to use:

Foo f = {10, 20.0};
std::cout << f;

which is very intuitive.

Answered   2023-09-20 20:30:22

  • I ve always read this regarding overloading << to be used with std::cout but what about overloading << to be used with the same class. In that case it can be member function right? - anyone

Comparison Operators, including Three-Way Comparison(C++20)

There are equality comparisons == and !=, and relational comparisons <, >, <=, >=. C++20 has also introduced the three-way comparison operator <=>.

Operator Meaning and Notes (Old) Meaning and Notes (C++20)
x == y true if x and y are equal

satisfies EqualityComparable
(used by std::unordered_map)
(x <=> y) == 0
(usually implemented directly, not
delegated to three-way unless = default)

satisfies std::equality_comparable
x != y !(x == y) !(x == y)
x < y true if x is lower than y

satisfies LessThanComparable
(used by std::set, std::sort, etc.
but requires strict weak ordering)
(x <=> y) < 0

may satisfy std::strict_weak_ordering
when wrapped in a functor
(e.g. std::ranges::less)
x > y y < x (x <=> y) > 0
x <= y !(x < y) for strong orderings,
x == y || x < y otherwise
(x <=> y) <= 0
x >= y y <= x (x <=> y) >= 0
x <=> y N/A three-way comparison
aka. "spaceship operator"

satisfies std::three_way_comparable

Guidelines

  1. Comparison operators shouldn't be member functions.1)
  2. If you define ==, define != too (unless it is rewritten in C++20).
  3. If you define <, define >, <=, and >= too.
  4. (C++20) Prefer defining <=> over defining each relational operator.
  5. (C++20) Prefer defaulting operators over implementing manually.
  6. Equality and relational comparisons should match, meaning that
    x == y should be equivalent to !(x < y) && !(y < x)2)
  7. Don't define == in terms of <, even when you could 3)

1) Otherwise, implicit conversions would be asymmetrical, and == is expected to apply the same kinds of implicit conversions to both sides.
2) This equivalence does not apply to float, but does apply to int and other strongly ordered types.
3) This is motivated by readability, correctness, and performance.

Implementation and Common Idioms Prior to C++20

Disclaimer
If you're using C++20, the implementations in this section have been obsoleted.
Skip ahead to the C++20 parts unless you're interested in a historical perspective.

All operators are typically implemented as non-member functions, possibly as hidden friends (friends where the function is defined inside the class). All following code examples use hidden friends because this becomes necessary if you need to compare private members anyway.

struct S {
    int x, y, z;

    // (In)equality comparison:
    // implementing a member-wise equality
    friend bool operator==(const S& l, const S& r) {
        return l.x == r.x && l.y == r.y && l.z == r.z;
    }
    friend bool operator!=(const S& l, const S& r) { return !(l == r); }

    // Relational comparisons:
    // implementing a lexicographical comparison which induces a
    // strict weak ordering.
    friend bool operator<(const S& l, const S& r) {
        if (l.x < r.x) return true;   // notice how all sub-comparisons
        if (r.x < l.x) return false;  // are implemented in terms of <
        if (l.y < r.y) return true;
        if (r.y < l.y) return false; // also see below for a possibly simpler
        return l.z < r.z;            // implementation
    }
    friend bool operator>(const S& l, const S& r) { return r < l; }
    friend bool operator<=(const S& l, const S& r) { return !(r < l); }
    friend bool operator>=(const S& l, const S& r) { return !(l < r); }
};

Note: in C++11, all of these can typically be noexcept and constexpr.

Implementing all relational comparisons in terms of < is not valid if we have a partially ordered member (e.g. float). In that case, <= and >= must be written differently.

friend bool operator<=(const S& l, const S& r) { return l == r || l < r; }
friend bool operator>=(const S& l, const S& r) { return r <= l; }

Further Notes on operator<

The implementation of operator< is not so simple because a proper lexicographical comparison cannot simply compare each member once. {1, 2} < {3, 0} should be true, even though 2 < 0 is false.

A lexicographical comparison is a simple way of implementing a strict weak ordering, which is needed for containers like std::set and algorithms like std::sort. In short, a strict weak ordering should behave like the < operator for integers, except that some integers are allowed to be equivalent (e.g. for all even integers, x < y is false).

If x != y is equivalent to x < y || y < x, a simpler approach is possible:

friend bool operator<(const S& l, const S& r) {
    if (l.x != r.x) return l.x < r.x;
    if (l.y != r.y) return l.y < r.y;
    return l.z < r.z;
}

Common Idioms

For multiple members, you can use std::tie to implement comparison lexicographically:

#include <tuple>

struct S {
    int x, y, z;

    friend bool operator<(const S& l, const S& r) {
        return std::tie(l.x, l.y, l.z) < std::tie(r.x, r.y, r.z);
    }
};

Use std::lexicographical_compare for array members.

Some people use macros or the curiously recurring template pattern (CRTP) to save the boilerplate of delegating !=, >, >=, and <=, or to imitate C++20's three-way comparison.

It is also possible to use std::rel_ops (deprecated in C++20) to delegate !=, >, <=, and >= to < and == for all types in some scope.


Default Comparisons (C++20)

A substantial amount of comparison operators simply compare each member of a class. If so, the implementation is pure boilerplate and we can let the compiler do it all:

struct S {
    int x, y, z;
    // ==, !=, <, >, <=, >= are all defined.
    // constexpr and noexcept are inferred automatically.
    friend auto operator<=>(const S&, const S&) = default;
};

Note: defaulted comparison operators need to be friends of the class, and the easiest way to accomplish that is by defining them as defaulted inside the class. This makes them "hidden friends".

Alternatively, we can default individual comparison operators. This is useful if we want to define equality comparison, or only relational comparison:

friend bool operator==(const S&, const S&) = default; // inside S

See the cppreference article on default comparison.

Expression Rewriting (C++20)

In C++20, if a comparison wasn't directly implemented, the compiler will also try to use rewrite candidates. Thanks to this, even if <=> isn't defaulted (which would implement all operators), we only have to implement == and <=>, and all other comparisons are rewritten in terms of these two.

Operator Potential Rewrites
x == y y == x
x != y !(x == y) or !(y == x) if equality comparison returns bool
x < y (x <=> y) < 0 or 0 < (y <=> x) if comparison result is comparable to zero
x > y (x <=> y) > 0 or 0 > (y <=> x) if ...
x <= y (x <=> y) <= 0 or 0 <= (y <=> x) if ...
x >= y (x <=> y) >= 0 or 0 >= (y <=> x) if ...
struct S {
    int x, y, z;
    // ==, !=
    friend constexpr bool operator==(const S& l, const S& r) noexcept { /* ... */ }
    // <=>, <, >, <=, >=
    friend constexpr auto operator<=>(const S& l, const S& r) noexcept { /* ... */ }
};

Note: constexpr and noexcept are optional, but can almost always be applied to comparison operators.

Three-Way Comparison Operator (C++20)

Note: it is colloquially called "spaceship operator". See also .

The basic idea behind x <=> y is that the result tells us whether x is lower than, greater than, equivalent to, or unordered with y. This is similar to functions like strcmp in C.

// old C style
int compare(int x, int y) {
    if (x < y) return -1;
    if (x > y) return  1;
    return             0; // or simply return (x > y) - (x < y);
}
// C++20 style: this is what <=> does for int.
auto compare_cxx20(int x, int y) {
    if (x < y) return std::strong_ordering::less;
    if (x > y) return std::strong_ordering::greater;
    return            std::strong_ordering::equal;
}
// This is what <=> does for float.
auto compare_cxx20(float x, float y) {
    if (x < y)  return std::partial_ordering::less;
    if (x > y)  return std::partial_ordering::greater;
    if (x == y) return std::partial_ordering::equivalent;
    return             std::partial_ordering::unordered; // NaN
}

Comparison Categories

The result of this operator is neither bool nor int, but a value of comparison category.

Comparison Category Example Possible Values
std::strong_ordering int less, equal = equivalent, greater
std::weak_ordering user-defined1) less, equivalent, greater
std::partial_ordering float less, equivalent, greater, unordered

std::strong_orderings can be converted to std::weak_ordering, which can be converted to std::partial_ordering. Values of these categories are comparable to (e.g. (x <=> y) == 0) and this has similar meaning to the compare function above. However, std::partial_ordering::unordered returns false for all comparisons.


1) There are no fundamental types for which x <=> y results in std::weak_ordering. Strong and weak orderings are interchangeable in practice; see Practical meaning of std::strong_ordering and std::weak_ordering.

Manual Implementation of Three-Way Comparison

Three-way comparison is often defaulted, but could be implemented manually like:

#include <compare> // necessary, even if we don't use std::is_eq

struct S {
    int x, y, z;
    // This implementation is the same as what the compiler would do
    // if we defaulted <=> with = default;
    friend constexpr auto operator<=>(const S& l, const S& r) noexcept {
        // C++17 if statement with declaration makes this more readable.
        // !std::is_eq(c) is not the same as std::is_neq(c); it is also true
        // for std::partial_order::unordered.
        if (auto c = l.x <=> r.x; !std::is_eq(c)) /* 1) */ return comp_x;
        if (auto c = l.y <=> r.y; !std::is_eq(c)) return comp_y;
        return l.y <=> r.y;
    }
    // == is not automatically defined in terms of <=>.
    friend constexpr bool operator==(const S&, const S&) = default;
};

If all members of S weren't the same type, then we could either specify the category explicitly (in the return type), or we could obtain it with std::common_comparison_category:

std::common_comparison_category_t<decltype(l.x <=> l.x), /* ... */>

1) Helper functions like std::is_neq compare the result of <=> to zero. They express intent more clearly, but you don't have to use them.

Common Idioms

Alternatively, we can let std::tie figure out the details:

#include <tuple>

struct S {
    int x, y, z;

    friend constexpr auto operator<=>(const S& l, const S& r) noexcept {
        return std::tie(l.x, l.y, l.z) <=> std::tie(r.x, r.y, r.z);
    }
};

Use std::lexicographical_compare_three_way for array members.

Answered   2023-09-20 20:30:22

Summary of canonical function signatures

Many operator overloads can return pretty much anything. For example, nothing stops you from returning void in operator==. However, only a few of these signatures are canonical, which means that you would normally write them that way, and that such an operator can be explicitly defaulted with = default.

Assignment operators

struct X {
  X& operator=(const X&) = default;     // copy assignment operator
  X& operator=(X&&) noexcept = default; // move assignment operator
};

Explicit defaulting with = default; is possible, but you can also implement assignment manually. Move assignment is almost always noexcept, although it isn't mandatory.

Comparison operators

#include <compare> // for comparison categories

struct X {
  friend auto operator<=>(const X&, const X&) = default; // defaulted three-way comparison
  friend std::strong_ordering<=>(const X&, const X&);    // manual three-way comparison

  friend bool operator==(const X&, const X&) = default;  // equality comparisons
  friend bool operator!=(const X&, const X&) = default;  // defaultable since C++20

  friend bool operator<(const X&, const X&) = default;   // relational comparisons
  friend bool operator>(const X&, const X&) = default;   // defaultable since C++20
  friend bool operator<=(const X&, const X&) = default;
  friend bool operator>=(const X&, const X&) = default;
};

See this answer for more information on when and how to default/implement comparisons.

Arithmetic operators

struct X {
  friend X operator+(const X&, const X&); // binary plus
  friend X operator*(const X&, const X&); // binary multiplication
  friend X operator-(const X&, const X&); // binary minus
  friend X operator/(const X&, const X&); // binary division
  friend X operator%(const X&, const X&); // binary remainder

  X operator+() const;                    // unary plus
  X operator-() const;                    // unary minus

  X& operator++();                        // prefix increment
  X& operator--();                        // prefix decrement
  X  operator++(int);                     // postfix decrement
  X  operator--(int);                     // postfix decrement

  X& operator+=(const X&);                // compound arithmetic assignment
  X& operator-=(const X&);
  X& operator*(const X&);
  X& operator/=(const X&);
  X& operator%=(const X&);
};

It is also possible to take the left operator of binary operators by value, but this is not recommended because it makes the signature asymmetric and inhibits compiler optimizations.

Bitwise operators

struct X {
  using difference_type = /* some integer type */;

  friend X operator&(const X&, const X&);         // bitwise AND
  friend X operator|(const X&, const X&);         // bitwise OR
  friend X operator^(const X&, const X&);         // bitwise XOR
  
  friend X operator<<(const X&, difference_type); // bitwise left-shift
  friend X operator>>(const X&, difference_type); // bitwise right-shift

  X operator~() const;                            // bitwise NOT

  X& operator&=(const X&);                        // compound bitwise assignment
  X& operator|=(const X&);
  X& operator^(const X&);
  X& operator/=(const X&);
  X& operator%=(const X&);
};

Stream insertion and extraction

#include <ostream> // std::ostream
#include <istream> // std::istream

struct X {
  friend std::ostream& operator<<(std::ostream&, const X&); // stream insertion
  friend std::istream& operator>>(std::istream&, X&);       // stream extraction
};

Function call operator

struct X {
  using result = /* ... */;

         result operator()(user-defined-args...) /* const / volatile / & / && */;
  static result operator()(user-defined-args...);           // since C++23
};

Subscript operator

struct X {
  using key_type = /* ... */;
  using value_type = /* ... */;

  const value_type& operator[](key_type) const;
        value_type& operator[](key_type);

  static value_type& operator[](key_type); // since C++23
};

Note that operator[] can accept multiple parameters since C++23.

Member access operators

struct X {
  using value_type = /* ... */;

  const value_type& operator*() const; // indirection operator
        value_type& operator*();

  const value_type* operator->() const; // arrow operator
        value_type* operator->();
};

Pointer-to-member operator

struct X {
  using member_type = /* ... */;
  using member_pointer_type = /* ... */;
  
  const member_type& operator->*(member_pointer_type) const;
        member_type& operator->*(member_pointer_type);
};

Address-of operator

struct X {
  using address_type = /* ... */;

  address_type operator&() const; // address-of operator
};

Logical operators

struct X {
  friend X operator&&(const X&, const X&); // logical AND
  friend X operator||(const X&, const X&); // logical OR
  friend X operator!(const X&);            // logical NOT
};

Note that these don't return bool because they only make sense if X is already a logical type that similar to bool.

User-defined conversions

struct X {
  using type = /* ... */;

  operator type() const;          // arbitrary implicit conversion

  explicit operator bool() const; // explicit/contextual conversion to bool

  template <typename T>
    requires /* ... */            // optionally constrained
  explicit operator T() const;    // conversion function template
};

Coroutine await

struct X {
  using awaiter = /* ... */;

  awaiter operator co_await() const;
};

Comma operator

struct X {
  using pair_type = /* ... */;

  // often a template to support combination of arbitrary types
  friend pair_type operator,(const X&, const X&);
};

Allocation functions

struct X {
  // class-specific allocation functions
  void* operator new(std::size_t);
  void* operator new[](std::size_t);
  void* operator new(std::size_t, std::align_val_t); // C++17
  void* operator new[](std::size_t, std::align_val_t); // C++17

  // class-specific placement allocation functions
  void* operator new(std::size_t, user-defined-args...);
  void* operator new[](std::size_t, user-defined-args...);
  void* operator new(std::size_t, std::align_val_t, user-defined-args...); // C++17
  void* operator new[](std::size_t, std::align_val_t, user-defined-args...); // C++17

  // class-specific usual deallocation functions
  void operator delete(void*);
  void operator delete[](void*);
  void operator delete(void*, std::align_val_t); // C++17
  void operator delete[](void*, std::align_val_t); // C++17
  void operator delete(void*, std::size_t);
  void operator delete[](void*, std::size_t);
  void operator delete(void*, std::size_t, std::align_val_t); // C++17
  void operator delete[](void*, std::size_t, std::align_val_t); // C++17

  // class-specific placement deallocation functions
  void operator delete(void*, user-defined-args...);
  void operator delete(void*, user-defined-args...);

  // class-specific usual destroying deallocation functions
  void operator delete(X*, std::destroying_delete_t); // C++20
  void operator delete(X*, std::destroying_delete_t, std::align_val_t); // C++20
  void operator delete(X*, std::destroying_delete_t, std::size_t); // C++20
  void operator delete(X*, std::destroying_delete_t, std::size_t, std::align_val_t); // C++20
};

// non-class specific replaceable allocation functions ...

void* operator new(std::size_t);
void* operator delete(void*);
// ...

Answered   2023-09-20 20:30:22

Making it short and simple, I'll be referring to some points, which I had come over the past week as I was learning Python and C++, oops and other things, so it goes as follows:

  1. The Arity of the operator can not be modified further than to what it is!

  2. Overloaded operators can only have one default argument which the function call operator rest it cannot.

  3. Only built in operator can be overloaded, rest can't!

For more info, you can refer to the following link, which redirects you to the documentation provided by GeekforGeeks.

https://www.geeksforgeeks.org/g-fact-39/

Answered   2023-09-20 20:30:22