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.)
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.
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;
}
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.
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.
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.
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.
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.
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.
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.
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 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 *
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 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 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.
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.
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.
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
boost::spirit
lol. - anyone +
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 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 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:
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
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 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 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
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.
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.
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
new
and delete
operatorsNote: 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.
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.
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.
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.
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
nothrow
new. - 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
<<
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 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 equalsatisfies 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 |
==
, define !=
too (unless it is rewritten in C++20).<
, define >
, <=
, and >=
too.<=>
over defining each relational operator.x == y
should be equivalent to !(x < y) && !(y < x)
2)==
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.
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 (friend
s 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; }
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;
}
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.
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 friend
s 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.
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.
Note: it is colloquially called "spaceship operator". See also spaceship-operator.
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
}
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_ordering
s 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.
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.
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
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
.
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.
#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.
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.
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&);
};
#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
};
struct X {
using result = /* ... */;
result operator()(user-defined-args...) /* const / volatile / & / && */;
static result operator()(user-defined-args...); // since C++23
};
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.
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->();
};
struct X {
using member_type = /* ... */;
using member_pointer_type = /* ... */;
const member_type& operator->*(member_pointer_type) const;
member_type& operator->*(member_pointer_type);
};
struct X {
using address_type = /* ... */;
address_type operator&() const; // address-of operator
};
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
.
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
};
struct X {
using awaiter = /* ... */;
awaiter operator co_await() const;
};
struct X {
using pair_type = /* ... */;
// often a template to support combination of arbitrary types
friend pair_type operator,(const X&, const X&);
};
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:
The Arity of the operator can not be modified further than to what it is!
Overloaded operators can only have one default argument which the function call operator rest it cannot.
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.
Answered 2023-09-20 20:30:22