Related
This is a question I've always pondered on and have never found any resource stating the answer to this question. In fact its not only for +=, but also for its siblings i.e. -=, *=, /=, etc. (of course not ==).
Consider the example,
int a = 5;
a += 4;
//this will make 'a' 9
Now consider the equivalent expression:
a = a + 4;
//This also makes 'a' 9
If += were simply a shorthand for a = a + <rhs of +=>
overloading + operator should also implicitly overload +=, unless explicitly overloaded otherwise. But that isn't what happens. That means, a += b doesn't get converted to a = a + b. But then why wasn't it implemented this way? As in, wouldn't it have been easier to simply convert it to a = a + b during compilation instead of implementing it separately as an operator in itself? That would also help in operator overloading, where a += b, where a and b are objects of the same class would not have to be explicitly overloaded, and simply overloading + would have been enough?
EDIT:
My question becomes more clear with this answer
Let me explain my question with an example where one needs to overload the operators:
class A {
int ivar;
public:
A() = default;
A(int par_ivar) : ivar(par_ivar) { }
A(A& a) {
this.ivar = a.ivar;
}
A(A&& a) noexcept {
this.ivar = a.ivar;
}
A operator+(const A& a) const {
A temp_a;
temp_a.ivar = this.ivar + a.ivar;
return temp_a;
}
void operator=(const A& a) {
this.ivar = a.ivar;
}
~A() = default;
};
Now, let's take a look at the result of 2 programs:
prog1:
int main() {
A a1(2);
A a2(3);
a1 = a1 + a2; //a1.ivar = 5
return 0;
}
prog2:
int main() {
A a1(2);
A a2(3);
a1 += a2; //compilation error!!
return 0;
}
Even when both the programs meant to do, nay, do the same thing, one compiles and runs (hoping that my overloads are correct) the other does not even compile!! Had += been simply replaced by appropriate + and =, we would not have felt the need for an explicit overload of +=. Was this intended to be, or is this a feature waiting to be added?
Operators are not generated from others (except with/from <=> in C++20):
providing operator < doesn't allow a > b (which is indeed "logically" equivalent to b < a). You have to implement all (even by re-using some).
For classes, a += b is not a shorthand for a = a + b
but for a.operator +=(b) or operator +=(a, b)
In the same way a = a + b is a shorthand for a.operator=(operator +(a, b)) (or its variant)
In practice, it is more efficient to implement operator+ from operator += than the reverse.
Even if a user might expect similar behavior according to their names, they are regular functions.
I already saw a matrix iterator for which ++it increase column index whereas it++ increase row index.
If += were simply a shorthand for a = a + <rhs of +=> overloading + operator should also implicitly overload +=, unless explicitly overloaded otherwise. But that isn't what happens. That means, a += b doesn't get converted to a = a + b.
(Possible) rational to not generate might be performance and control:
Vector (for math) or Matrix are good example:
4 possible overloads
Matrix operator+(Matrix&& lhs, Matrix&& rhs) { return std::move(lhs += rhs); }
Matrix operator+(Matrix&& lhs, const Matrix& rhs) { return std::move(lhs += rhs); }
Matrix operator+(const Matrix& lhs, Matrix&& rhs) { return std::move(rhs += lhs); } // + is symmetrical :)
Matrix operator+(const Matrix& lhs, const Matrix& rhs) { auto tmp{lhs}; return tmp += rhs; }
Side effect of the decision allow to give different meanings to operators, as "name operator":
if (42 <in> std::vector{4, 8, 15, 16, 23, 42})
Using a = a + b will imply using a copy assignment (as operator = is used). On the other hand, a += b is by default a compound assignment.
According to cppreference,
copy assignment operator replaces the contents of the object a with a copy of the contents of b (b is not modified).
and
compound assignment operators replace the contents of the object a with the result of a binary operation between the previous value of a and the value of b.
Using a = a + b, would therefore cause an unnecessary memory usage, since a has to be copied once before its value is changed.
MATLAB arrays support matrix operations and element operations. For example, M*N and M.*N. This is a quite intuitive way to distinguish the two different operations. If I want to implement similar operations in C++, how can I do that?
Can I create a new operator, .*, too? If yes, can anyone give me some guidance?
No, you can't overload op.*:
[C++03 & C++11: 13.5/3]: The following operators cannot be overloaded:
. .* :: ?:
In C++, there's a list of predefined operators, most of which are overloadable (.* is not). Additionally, any name can be used as an operator like:
#include <iostream>
// generic LHSlt holder
template<typename LHS, typename OP>
struct LHSlt {
LHS lhs_;
};
// declare myop as an operator-like construct
enum { myop };
// parse 'lhs <myop' into LHSlt
template<typename LHS>
LHSlt<LHS, decltype(myop)> operator<(const LHS& lhs, decltype(myop))
{
return { lhs };
}
// declare (int <myop> int) -> int
int operator>(LHSlt<int, decltype(myop)> lhsof, int rhs)
{
int& lhs = lhsof.lhs_;
// here comes your actual implementation
return (lhs + rhs) * (lhs - rhs);
}
// strictly optional
#define MYOP <myop>
int main() {
std::cout << (5 <myop> 2) << ' ' << (5 MYOP 2);
}
Disclaimer: This, strictly speaking, gets translated to (5 < myop) > 2, which is LHSlt<int, decltype(myop)>(5) > 2. Thus it's not a new 'operator', in C++-terms, but it's used exactly the same way, even in terms of ADL. Also, if type is large, you probably want to store const T&.
Note that you can do this with any binary operator that can be defined external to the class; precedence is based on the precedence of the two sides (< and >). Thus you can have e.g. *myop*, +myop+, <<myop>>, <myop>, |myop| in this order of precedence.
If you want right-associativity, it gets a bit more tricky. You'll need both of a RHS-holder and LHS-holder (the latter being LHSlt here) and use surrounding operators such that the right one has higher precedence than the left one, e.g. a |myop> b |myop>c is a |myop> (b |myop> c). Then you need the function for both your type and your holder type as the lhs.
You cannot overload .* (see Lightness' answer for standard text), but, interestingly enough, you can overload ->* (similar to how you can overload -> but not .). If that's sufficient for differentiation, then have at it:
struct Int {
int i;
Int operator*(Int rhs) const { return Int{i * rhs.i}; }
Int operator->*(Int rhs) const { return Int{i + rhs.i}; }
friend std::ostream& operator<<(std::ostream& os, Int rhs) {
return os << "Int(" << rhs.i << ')';
}
};
int main() {
Int five{5};
Int six{6};
std::cout << (five * six) << ", " << (five ->* six) << '\n';
}
That'll print Int(30), Int(11).
No, unfortunately you cannot define new operators—you can only overload existing operators (with a few important exceptions, such as operator.). Even then, it's typically only a good idea to overload operators for types which have very clear and uncontroversial existing semantics for a given operator—for instance, any type that behaves as a number is a good candidate for overloading the arithmetic and comparison operators, but you should make sure that operator+ doesn't, say, subtract two numbers.
MATLAB arrays support matrix operations and element operations. For example, M*N and M.*N. This is a quite intuitive way to distinguish the two different operations. If I want to implement similar operations in C++, how can I do that?
Can I create a new operator, .*, too? If yes, can anyone give me some guidance?
As for the first part you can overload most of the operators and there are some that you can not overload and the list of operators in C++ are:
Arithmetic
+ (addition)
- (subtraction)
* (multiplication)
/ (division)
% (modulus)
Bitwise
^ (XOR)
| (OR)
& (AND)
~ (Complement)
<< (Shift Left, Insertion to Stream)
>> (Shift Right, Extraction from Stream)
Assignment
= (Assignment)
Relational
== (Equality)
!= (Inequality)
> (Greater-Than)
< (Less-Than)
>= (Greater-Than Or Equal-To)
<= (Less-Than Or Equal-To)
Logical
! (NOT)
&& (AND)
|| (OR)
Compound Assignment
+= (Addition-Assignment)
-= (Subtraction-Assignment)
*= (Multiplication-Assignment)
/= (Division-Assignment)
%= (Modulus-Assignment)
&= (AND-Assignment)
|= (OR-Assignment)
^= (XOR-Assignment)
<<= (Shift-Left Assignment)
>>= (Shift-Right Assignment)
Increment - Decrement - Both have 2 forms (prefix) and (postfix)
++ (Increment)
-- (Decrement)
Subscript
[] (Subscript)
Function Call
() (Function Call)
Address, Reference, Pointer
operator&()
operator*()
operator->()
Comma
operator,()
Member Reference
operator->()
operator->*()
Memory Management
new
delete
new[]
delete[]
Conversion
operator "type" () const
NON Modifiable Operators - Operators that can not be overloaded
?: (Conditional - Ternary)
. (Member Selection)
.* (Member Selection With Pointer To Member)
:: (Scope Resolution)
sizeof() (Object Size Information)
typeid() (Object Type Information)
So knowing this list will help to answer your questions. Can you Create a "New Operator" in C++? No! If you want to implement similar operations in C++; how can I do that?
You have 4 choices: Either overload an already existing operator that can be overloaded, write a function or method to do the type of calculations you want to perform, create a template type to do the work for you, or the last one which is the least common to do but you can also write macros to do them for you.
There is a header only Math API Library that is used quite frequently with OpenGL graphics API and OpenGL's Shader Language GLSL and this library has many features that work with vectors, matrices, quaternions etc., and all the necessary functions and operations that can be done to them. Here is the link to GLM You can have a look at their documentation as well as their library implementations since it is a headers only library or API. This should give you some insight on how they constructed their Vector and Matrix objects and the operations that can be done to them.
BTW: I am seeking to answer the parts of this question as asked. I am also not seeking to replicate all the information in other worthy answers. The bounty seeks something different to the question as asked, so I am not responding to that.
It is actually fairly simple to provide a matrix multiplication. Since I'm not proposing to describe data structures to represent a matrix and fully implement operations and validity checks on them, I'll just provide skeletons to illustrate.
Example 1: operator*() as a member function
class M // a basic matrix class
{
public:
// assume other constructors and members to set things up
M operator*(const M &rhs) const;
};
M M::operator*(const M &rhs) const
{
// implement checks on dimensions, throw an exception if invalid
M result;
// implement the multiplication (typical iterations) and store results in result
return result;
}
int main()
{
M a;
M b;
// set up elements of a and b as needed
M c = a*b; // this relies on M having appropriate constructor(s) to copy or move the result of a*b into c
M d;
d = a * b; // this relies on M having appropriate operator=() to assign d to the result of a*b
}
The above implements operator*() as a member function. So, functionally, c = a*b is equivalent to c = a.operator*(b). The const qualifiers represent the fact that a matrix multiplication a*b does not generally change a or b.
Example 2: operator*() as a non-member function
Now, operator*() can also be implemented as a non-member (optionally a friend), with a skeleton that looks like
class M // our basic matrix class, different operator *
{
public:
// assume other constructors and members to set things up
friend M operator*(const M &lhs, const M &rhs);
};
M operator*(const M &lhs, const M &rhs)
{
// implement checks on dimensions, throw an exception if invalid
M result;
// implement the multiplication (typical iterations) and store results in result
return result;
}
// same main() as before
Note that, in this case, a*b is now equivalent to operator*(a, b).
If you want to use both forms, care is needed to avoid ambiguity. If both forms of operator*() are provided they are both valid matches in a statement like c = a*b and the compiler has no means to choose one form over the other. The result is code not compiling.
Example 3: overloading operator*()
It is also possible to overload operator*() - for example, to multiply a matrix by a scalar.
class M // a basic matrix class
{
public:
// assume other constructors and members to set things up
M operator*(const M &rhs) const; // as in first example
M operator*(double scalar) const; // member form
friend M operator*(double scalar, const M &rhs); // non-member form
};
M M::operator*(double scalar) const
{
M result;
// implement the multiplication (typical iterations) and store results in result
return result;
}
M operator*(double scalar, const M &m)
{
M result;
// implement the multiplication (typical iterations) and store results in result
return result;
}
int main()
{
M a;
M b;
// set up elements of a and b as needed
M c = b * 2.0; // uses the member form of operator*() above
M d;
d = 2.0*a; // uses the non-member form of operator*() above
}
In the above b*2.0 amounts to a call of b.operator*(2.0) and 2.0*a to a call of the non-member operator*(2.0, a). The member forms can only generally be used in expressions where the left hand operand is of type M. So 2.0*a will not work if only member forms of operator*() is provided.
Discussion
Apart from concerns of ambiguity above, there are other things to be aware of when overloading operators.
It is not possible to change precedence or associativity of operators from their specification in language rules. So, in the expression a+b*c, the * will always have higher precedence than the +. This is also the reason it is not a good idea to overload ^ for exponentiation in C++, since ^ has a lower precedence than + in C++ (being a bitwise operation on integral types). So a + b^c is actually equivalent in C++ to (a + b)^c, not to a + (b^c) (which anyone with basic knowledge of algebra would expect).
The language specifies a set of operators, and it is not possible to create new ones. For example, there is no ** in C++, such that a ** b raises a to the power of b (which other languages can do), and it is not possible to create one.
Not all operators can be overloaded.
One of the operators that cannot be overloaded in C++ is .*. So it is not possible to use such an operator like you would in Matlab. I would generally suggest NOT trying to get the same effect using other operators, because the above constraints will affect that (and cause expressions to give counter-intuitive behaviour). Instead simply provide another named function to do the job. For example, as a member function
class M
{
public:
// other stuff
M ElementWiseProduct(const M &) const;
};
Most of the answers have already covered which operators are and are not overloadable, but none have discussed WHY some are mutable and some aren't.
The following is a quote from Bjarne Stroustrup (the guy who wrote c++) that I found in this stackoverflow answer. Pay particular attention to the third paragraph.
When I decided to allow overloading of operator ->, I naturally considered whether operator . could be similarly overloaded.
At the time, I considered the following arguments conclusive: If obj is a class object then obj.m has a meaning for every member m of that object's class. We try not to make the language mutable by redefining built-in operations (though that rule is violated for = out of dire need, and for unary &).
If we allowed overloading of . for a class X, we would be unable to access members of X by normal means; we would have to use a pointer and ->, but -> and & might also have been re-defined. I wanted an extensible language, not a mutable one.
These arguments are weighty, but not conclusive. In particular, in 1990 Jim Adcock proposed to allow overloading of operator . exactly the way operator -> is.
A page on his website adds a little more:
Can I define my own operators?
Sorry, no. The possibility has been considered several times, but each time I/we decided that the likely problems outweighed the likely benefits.
It's not a language-technical problem. Even when I first considerd it in 1983, I knew how it could be implemented. However, my experience has been that when we go beyond the most trivial examples people seem to have subtlely different opinions of "the obvious" meaning of uses of an operator. A classical example is a ** b ** c. Assume that ** has been made to mean exponentiation. Now should a ** b ** c mean (a ** b) ** c or a ** (b ** c)? I thought the answer was obvious and my friends agreed - and then we found that we didn't agree on which resolution was the obvious one. My conjecture is that such problems would lead to subtle bugs.
So, while most operators can be overloaded, it was never intended for people to create arbitrary operators in c++.
It's as simple (and as difficult!) as defining a function named (in this case) operator*():
Matrix operator*(const Matrix &m1, const Matrix &m2) ...
where Matrix is a class you've defined to represent matrices.
As other answers say, overloading operator.* is not possible.
But I got a good solution for your question, check here.
You can provide any methods in operator-ish form like:
M <matrix_mul> N
As I read in books and in the web, in C++ we can overload the "plus" or "minus" operators with these prototypes (as member functions of a class Money):
const Money operator +(const Money& m2) const;
const Money operator -(const Money& m2) const;
and for the assignment operator with:
const Money& operator =(const Money& m2);
Why use a reference to a Money object as a return value in the assignment operator overloading and not in the plus and minus operators?
Returning a reference from assignment allows chaining:
a = b = c; // shorter than the equivalent "b = c; a = b;"
(This would also work (in most cases) if the operator returned a copy of the new value, but that's generally less efficient.)
We can't return a reference from arithmetic operations, since they produce a new value. The only (sensible) way to return a new value is to return it by value.
Returning a constant value, as your example does, prevents move semantics, so don't do that.
Because operator+ and operator- don't act on this object, but return a new object that is the summation (or subtraction) of this object from another.
operator= is different because it's actually assigning something to this object.
operator+= and operator-= would act on this object, and are a closer analog to operator=.
Consider what you are asking. You would want an expression, a + b, to return a reference to one of a or b, which would have the results of the expression. Thus you would modify one of a or b to be the sum of a and b. So you would want to redefine the semantics of the operator (+) to be the same as the operator (+=). And like #manuell said, you would thus allow (a + b) = c. The semantics you are suggesting are already offered by += and -=.
I think its fine if you return by value in overloaded assignment operator , that is because of associativity of assignment operator. consider this:
int a = b =c = 3 ;
here associativity is as followed:
(a=(b=(c=3)))
but consider iostream operation
cout << x << y << z ;
here associativity is as followed:
(((cout << x )<< y) << z) ;
you can see that x will be printed first , so if you return by value in overloading of << operator , return value will not be "lvalue" , while returning by refrence is a lvalue , so cascading of << operator can be achieve.
one more point , copy constructor will get called if you return by value. ( which is not the case with return by refrence)
The link shown below has better explanation I guess
return value of operator overloading in C++
Apart from efficiency during chained assignment (e.g., a = b = c;) as pointed out by the accepted answer, it is also a matter of convention. Please, refer Item 10 of Effective C++ by Scott Meyers. It is a convention followed by the built-in types and all the types in the standard library.
MATLAB arrays support matrix operations and element operations. For example, M*N and M.*N. This is a quite intuitive way to distinguish the two different operations. If I want to implement similar operations in C++, how can I do that?
Can I create a new operator, .*, too? If yes, can anyone give me some guidance?
No, you can't overload op.*:
[C++03 & C++11: 13.5/3]: The following operators cannot be overloaded:
. .* :: ?:
In C++, there's a list of predefined operators, most of which are overloadable (.* is not). Additionally, any name can be used as an operator like:
#include <iostream>
// generic LHSlt holder
template<typename LHS, typename OP>
struct LHSlt {
LHS lhs_;
};
// declare myop as an operator-like construct
enum { myop };
// parse 'lhs <myop' into LHSlt
template<typename LHS>
LHSlt<LHS, decltype(myop)> operator<(const LHS& lhs, decltype(myop))
{
return { lhs };
}
// declare (int <myop> int) -> int
int operator>(LHSlt<int, decltype(myop)> lhsof, int rhs)
{
int& lhs = lhsof.lhs_;
// here comes your actual implementation
return (lhs + rhs) * (lhs - rhs);
}
// strictly optional
#define MYOP <myop>
int main() {
std::cout << (5 <myop> 2) << ' ' << (5 MYOP 2);
}
Disclaimer: This, strictly speaking, gets translated to (5 < myop) > 2, which is LHSlt<int, decltype(myop)>(5) > 2. Thus it's not a new 'operator', in C++-terms, but it's used exactly the same way, even in terms of ADL. Also, if type is large, you probably want to store const T&.
Note that you can do this with any binary operator that can be defined external to the class; precedence is based on the precedence of the two sides (< and >). Thus you can have e.g. *myop*, +myop+, <<myop>>, <myop>, |myop| in this order of precedence.
If you want right-associativity, it gets a bit more tricky. You'll need both of a RHS-holder and LHS-holder (the latter being LHSlt here) and use surrounding operators such that the right one has higher precedence than the left one, e.g. a |myop> b |myop>c is a |myop> (b |myop> c). Then you need the function for both your type and your holder type as the lhs.
You cannot overload .* (see Lightness' answer for standard text), but, interestingly enough, you can overload ->* (similar to how you can overload -> but not .). If that's sufficient for differentiation, then have at it:
struct Int {
int i;
Int operator*(Int rhs) const { return Int{i * rhs.i}; }
Int operator->*(Int rhs) const { return Int{i + rhs.i}; }
friend std::ostream& operator<<(std::ostream& os, Int rhs) {
return os << "Int(" << rhs.i << ')';
}
};
int main() {
Int five{5};
Int six{6};
std::cout << (five * six) << ", " << (five ->* six) << '\n';
}
That'll print Int(30), Int(11).
No, unfortunately you cannot define new operators—you can only overload existing operators (with a few important exceptions, such as operator.). Even then, it's typically only a good idea to overload operators for types which have very clear and uncontroversial existing semantics for a given operator—for instance, any type that behaves as a number is a good candidate for overloading the arithmetic and comparison operators, but you should make sure that operator+ doesn't, say, subtract two numbers.
MATLAB arrays support matrix operations and element operations. For example, M*N and M.*N. This is a quite intuitive way to distinguish the two different operations. If I want to implement similar operations in C++, how can I do that?
Can I create a new operator, .*, too? If yes, can anyone give me some guidance?
As for the first part you can overload most of the operators and there are some that you can not overload and the list of operators in C++ are:
Arithmetic
+ (addition)
- (subtraction)
* (multiplication)
/ (division)
% (modulus)
Bitwise
^ (XOR)
| (OR)
& (AND)
~ (Complement)
<< (Shift Left, Insertion to Stream)
>> (Shift Right, Extraction from Stream)
Assignment
= (Assignment)
Relational
== (Equality)
!= (Inequality)
> (Greater-Than)
< (Less-Than)
>= (Greater-Than Or Equal-To)
<= (Less-Than Or Equal-To)
Logical
! (NOT)
&& (AND)
|| (OR)
Compound Assignment
+= (Addition-Assignment)
-= (Subtraction-Assignment)
*= (Multiplication-Assignment)
/= (Division-Assignment)
%= (Modulus-Assignment)
&= (AND-Assignment)
|= (OR-Assignment)
^= (XOR-Assignment)
<<= (Shift-Left Assignment)
>>= (Shift-Right Assignment)
Increment - Decrement - Both have 2 forms (prefix) and (postfix)
++ (Increment)
-- (Decrement)
Subscript
[] (Subscript)
Function Call
() (Function Call)
Address, Reference, Pointer
operator&()
operator*()
operator->()
Comma
operator,()
Member Reference
operator->()
operator->*()
Memory Management
new
delete
new[]
delete[]
Conversion
operator "type" () const
NON Modifiable Operators - Operators that can not be overloaded
?: (Conditional - Ternary)
. (Member Selection)
.* (Member Selection With Pointer To Member)
:: (Scope Resolution)
sizeof() (Object Size Information)
typeid() (Object Type Information)
So knowing this list will help to answer your questions. Can you Create a "New Operator" in C++? No! If you want to implement similar operations in C++; how can I do that?
You have 4 choices: Either overload an already existing operator that can be overloaded, write a function or method to do the type of calculations you want to perform, create a template type to do the work for you, or the last one which is the least common to do but you can also write macros to do them for you.
There is a header only Math API Library that is used quite frequently with OpenGL graphics API and OpenGL's Shader Language GLSL and this library has many features that work with vectors, matrices, quaternions etc., and all the necessary functions and operations that can be done to them. Here is the link to GLM You can have a look at their documentation as well as their library implementations since it is a headers only library or API. This should give you some insight on how they constructed their Vector and Matrix objects and the operations that can be done to them.
BTW: I am seeking to answer the parts of this question as asked. I am also not seeking to replicate all the information in other worthy answers. The bounty seeks something different to the question as asked, so I am not responding to that.
It is actually fairly simple to provide a matrix multiplication. Since I'm not proposing to describe data structures to represent a matrix and fully implement operations and validity checks on them, I'll just provide skeletons to illustrate.
Example 1: operator*() as a member function
class M // a basic matrix class
{
public:
// assume other constructors and members to set things up
M operator*(const M &rhs) const;
};
M M::operator*(const M &rhs) const
{
// implement checks on dimensions, throw an exception if invalid
M result;
// implement the multiplication (typical iterations) and store results in result
return result;
}
int main()
{
M a;
M b;
// set up elements of a and b as needed
M c = a*b; // this relies on M having appropriate constructor(s) to copy or move the result of a*b into c
M d;
d = a * b; // this relies on M having appropriate operator=() to assign d to the result of a*b
}
The above implements operator*() as a member function. So, functionally, c = a*b is equivalent to c = a.operator*(b). The const qualifiers represent the fact that a matrix multiplication a*b does not generally change a or b.
Example 2: operator*() as a non-member function
Now, operator*() can also be implemented as a non-member (optionally a friend), with a skeleton that looks like
class M // our basic matrix class, different operator *
{
public:
// assume other constructors and members to set things up
friend M operator*(const M &lhs, const M &rhs);
};
M operator*(const M &lhs, const M &rhs)
{
// implement checks on dimensions, throw an exception if invalid
M result;
// implement the multiplication (typical iterations) and store results in result
return result;
}
// same main() as before
Note that, in this case, a*b is now equivalent to operator*(a, b).
If you want to use both forms, care is needed to avoid ambiguity. If both forms of operator*() are provided they are both valid matches in a statement like c = a*b and the compiler has no means to choose one form over the other. The result is code not compiling.
Example 3: overloading operator*()
It is also possible to overload operator*() - for example, to multiply a matrix by a scalar.
class M // a basic matrix class
{
public:
// assume other constructors and members to set things up
M operator*(const M &rhs) const; // as in first example
M operator*(double scalar) const; // member form
friend M operator*(double scalar, const M &rhs); // non-member form
};
M M::operator*(double scalar) const
{
M result;
// implement the multiplication (typical iterations) and store results in result
return result;
}
M operator*(double scalar, const M &m)
{
M result;
// implement the multiplication (typical iterations) and store results in result
return result;
}
int main()
{
M a;
M b;
// set up elements of a and b as needed
M c = b * 2.0; // uses the member form of operator*() above
M d;
d = 2.0*a; // uses the non-member form of operator*() above
}
In the above b*2.0 amounts to a call of b.operator*(2.0) and 2.0*a to a call of the non-member operator*(2.0, a). The member forms can only generally be used in expressions where the left hand operand is of type M. So 2.0*a will not work if only member forms of operator*() is provided.
Discussion
Apart from concerns of ambiguity above, there are other things to be aware of when overloading operators.
It is not possible to change precedence or associativity of operators from their specification in language rules. So, in the expression a+b*c, the * will always have higher precedence than the +. This is also the reason it is not a good idea to overload ^ for exponentiation in C++, since ^ has a lower precedence than + in C++ (being a bitwise operation on integral types). So a + b^c is actually equivalent in C++ to (a + b)^c, not to a + (b^c) (which anyone with basic knowledge of algebra would expect).
The language specifies a set of operators, and it is not possible to create new ones. For example, there is no ** in C++, such that a ** b raises a to the power of b (which other languages can do), and it is not possible to create one.
Not all operators can be overloaded.
One of the operators that cannot be overloaded in C++ is .*. So it is not possible to use such an operator like you would in Matlab. I would generally suggest NOT trying to get the same effect using other operators, because the above constraints will affect that (and cause expressions to give counter-intuitive behaviour). Instead simply provide another named function to do the job. For example, as a member function
class M
{
public:
// other stuff
M ElementWiseProduct(const M &) const;
};
Most of the answers have already covered which operators are and are not overloadable, but none have discussed WHY some are mutable and some aren't.
The following is a quote from Bjarne Stroustrup (the guy who wrote c++) that I found in this stackoverflow answer. Pay particular attention to the third paragraph.
When I decided to allow overloading of operator ->, I naturally considered whether operator . could be similarly overloaded.
At the time, I considered the following arguments conclusive: If obj is a class object then obj.m has a meaning for every member m of that object's class. We try not to make the language mutable by redefining built-in operations (though that rule is violated for = out of dire need, and for unary &).
If we allowed overloading of . for a class X, we would be unable to access members of X by normal means; we would have to use a pointer and ->, but -> and & might also have been re-defined. I wanted an extensible language, not a mutable one.
These arguments are weighty, but not conclusive. In particular, in 1990 Jim Adcock proposed to allow overloading of operator . exactly the way operator -> is.
A page on his website adds a little more:
Can I define my own operators?
Sorry, no. The possibility has been considered several times, but each time I/we decided that the likely problems outweighed the likely benefits.
It's not a language-technical problem. Even when I first considerd it in 1983, I knew how it could be implemented. However, my experience has been that when we go beyond the most trivial examples people seem to have subtlely different opinions of "the obvious" meaning of uses of an operator. A classical example is a ** b ** c. Assume that ** has been made to mean exponentiation. Now should a ** b ** c mean (a ** b) ** c or a ** (b ** c)? I thought the answer was obvious and my friends agreed - and then we found that we didn't agree on which resolution was the obvious one. My conjecture is that such problems would lead to subtle bugs.
So, while most operators can be overloaded, it was never intended for people to create arbitrary operators in c++.
It's as simple (and as difficult!) as defining a function named (in this case) operator*():
Matrix operator*(const Matrix &m1, const Matrix &m2) ...
where Matrix is a class you've defined to represent matrices.
As other answers say, overloading operator.* is not possible.
But I got a good solution for your question, check here.
You can provide any methods in operator-ish form like:
M <matrix_mul> N
I realize this is a basic question but I have searched online, been to cplusplus.com, read through my book, and I can't seem to grasp the concept of overloaded operators. A specific example from cplusplus.com is:
// vectors: overloading operators example
#include <iostream>
using namespace std;
class CVector {
public:
int x,y;
CVector () {};
CVector (int,int);
CVector operator + (CVector);
};
CVector::CVector (int a, int b) {
x = a;
y = b;
}
CVector CVector::operator+ (CVector param) {
CVector temp;
temp.x = x + param.x;
temp.y = y + param.y;
return (temp);
}
int main () {
CVector a (3,1);
CVector b (1,2);
CVector c;
c = a + b;
cout << c.x << "," << c.y;
return 0;
}
From http://www.cplusplus.com/doc/tutorial/classes2/ but reading through it I'm still not understanding them at all. I just need a basic example of the point of the overloaded operator (which I assume is the "CVector CVector::operator+ (CVector param)").
There's also this example from wikipedia:
Time operator+(const Time& lhs, const Time& rhs)
{
Time temp = lhs;
temp.seconds += rhs.seconds;
if (temp.seconds >= 60)
{
temp.seconds -= 60;
temp.minutes++;
}
temp.minutes += rhs.minutes;
if (temp.minutes >= 60)
{
temp.minutes -= 60;
temp.hours++;
}
temp.hours += rhs.hours;
return temp;
}
From "http://en.wikipedia.org/wiki/Operator_overloading"
The current assignment I'm working on I need to overload a ++ and a -- operator.
Thanks in advance for the information and sorry about the somewhat vague question, unfortunately I'm just not sure on it at all.
Operator overloading is the technique that C++ provides to let you define how the operators in the language can be applied to non-built in objects.
In you example for the Time class operator overload for the + operator:
Time operator+(const Time& lhs, const Time& rhs);
With that overload, you can now perform addition operations on Time objects in a 'natural' fashion:
Time t1 = some_time_initializer;
Time t2 = some_other_time_initializer;
Time t3 = t1 + t2; // calls operator+( t1, t2)
The overload for an operator is just a function with the special name "operator" followed by the symbol for the operator being overloaded. Most operators can be overloaded - ones that cannot are:
. .* :: and ?:
You can call the function directly by name, but usually don't (the point of operator overloading is to be able to use the operators normally).
The overloaded function that gets called is determined by normal overload resolution on the arguments to the operator - that's how the compiler knows to call the operator+() that uses the Time argument types from the example above.
One additional thing to be aware of when overloading the ++ and -- increment and decrement operators is that there are two versions of each - the prefix and the postfix forms. The postfix version of these operators takes an extra int parameter (which is passed 0 and has no purpose other than to differentiate between the two types of operator). The C++ standard has the following examples:
class X {
public:
X& operator++(); //prefix ++a
X operator++(int); //postfix a++
};
class Y { };
Y& operator++(Y&); //prefix ++b
Y operator++(Y&, int); //postfix b++
You should also be aware that the overloaded operators do not have to perform operations that are similar to the built in operators - being more or less normal functions they can do whatever you want. For example, the standard library's IO stream interface uses the shift operators for output and input to/from streams - which is really nothing like bit shifting. However, if you try to be too fancy with your operator overloads, you'll cause much confusion for people who try to follow your code (maybe even you when you look at your code later).
Use operator overloading with care.
An operator in C++ is just a function with a special name. So instead of saying Add(int,int) you say operator +(int,int).
Now as any other function, you can overload it to say work on other types. In your vector example, if you overload operator + to take CVector arguments (ie. operator +(CVector, CVector)), you can then say:
CVector a,b,res;
res=a+b;
Since ++ and -- are unary (they take only one argument), to overload them you'd do like:
type operator ++(type p)
{
type res;
res.value++;
return res;
}
Where type is any type that has a field called value. You get the idea.
What you found in those references are not bad examples of when you'd want operator overloading (giving meaning to vector addition, for example), but they're horrible code when it comes down to the details.
For example, this is much more realistic, showing delegating to the compound assignment operator and proper marking of a const member function:
class Vector2
{
double m_x, m_y;
public:
Vector2(double x, double y) : m_x(x), m_y(y) {}
// Vector2(const Vector2& other) = default;
// Vector2& operator=(const Vector2& other) = default;
Vector2& operator+=(const Vector2& addend) { m_x += addend.m_x; m_y += addend.m_y; return *this; }
Vector2 operator+(const Vector2& addend) const { Vector2 sum(*this); return sum += addend; }
};
From your comments above, you dont see the point of all this operator overloading?
Operator overloading is simply 'syntactic sugar' hiding a method call, and making code somehwhat clearer in many cases.
Consider a simple Integer class wrapping an int. You would write add and other arithmetic methods, possibly increment and decrement as well, requiring a method call such as my_int.add(5). now renaming the add method to operator+ allows my_int + 5, which is more intuitive and clearer, cleaner code. But all it is really doing is hiding a call to your operator+ (renamed add?) method.
Things do get a bit more complex though, as operator + for numbers is well understood by everyone above 2nd grade. But as in the string example above, operators should usually only be applied where they have an intuitive meaning. The Apples example is a good example of where NOT to overload operators.
But applied to say, a List class, something like myList + anObject, should be intuitively understood as 'add anObject to myList', hence the use of the + operator. And operator '-' as meaning 'Removal from the list'.
As I said above, the point of all this is to make code (hopefully) clearer, as in the List example, which would you rather code? (and which do you find easier to read?) myList.add( anObject ) or myList + onObject? But in the background, a method (your implementation of operator+, or add) is being called either way. You can almost think of the compiler rewritting the code: my_int + 5 would become my_int.operator+(5)
All the examples given, such as Time and Vector classes, all have intuitive definitions for the operators. Vector addition... again, easier to code (and read) v1 = v2 + v3 than v1 = v2.add(v3). This is where all the caution you are likely to read regarding not going overboard with operators in your classes, because for most they just wont make sense. But of course there is nothing stopping you putting an operator & into a class like Apple, just dont expect others to know what it does without seeing the code for it!
'Overloading' the operator simply means your are supplying the compiler with another definition for that operator, applied to instances of your class. Rather like overloading methods, same name... different parameters...
Hope this helps...
The "operator" in this case is the + symbol.
The idea here is that an operator does something. An overloaded operator does something different.
So, in this case, the '+' operator, normally used to add two numbers, is being "overloaded" to allow for adding vectors or time.
EDIT: Adding two integers is built-in to c++; the compiler automatically understands what you mean when you do
int x, y = 2, z = 2;
x = y + z;
Objects, on the other hand, can be anything, so using a '+' between two objects doesn't inherently make any sense. If you have something like
Apple apple1, apple2, apple3;
apple3 = apple1 + apple2;
What does it mean when you add two Apple objects together? Nothing, until you overload the '+' operator and tell the compiler what it is that you mean when you add two Apple objects together.
An overloaded operator is when you use an operator to work with types that C++ doesn't "natively" support for that operator.
For example, you can typically use the binary "+" operator to add numeric values (floats, ints, doubles, etc.). You can also add an integer type to a pointer - for instance:
char foo[] = "A few words";
char *p = &(foo[3]); // Points to "e"
char *q = foo + 3; // Also points to "e"
But that's it! You can't do any more natively with a binary "+" operator.
However, operator overloading lets you do things the designers of C++ didn't build into the language - like use the + operator to concatenate strings - for instance:
std::string a("A short"), b(" string.");
std::string c = a + b; // c is "A short string."
Once you wrap your head around that, the Wikipedia examples will make more sense.
A operator would be "+", "-" or "+=". These perform different methods on existing objects. This in fact comes down to a method call. Other than normal method calls these look much more natural to a human user. Writing "1 + 2" just looks more normal and is shorter than "add(1,2)". If you overload an operator, you change the method it executes.
In your first example, the "+" operator's method is overloaded, so that you can use it for vector-addition.
I would suggest that you copy the first example into an editor and play a little around with it. Once you understand what the code does, my suggestion would be to implement vector subtraction and multiplication.
Before starting out, there are many operators out there! Here is a list of all C++ operators: list.
With this being said, operator overloading in C++ is a way to make a certain operator behave in a particular way for an object.
For example, if you use the increment/decrement operators (++ and --) on an object, the compiler will not understand what needs to be incremented/decremented in the object because it is not a primitive type (int, char, float...). You must define the appropriate behavior for the compiler to understand what you mean. Operator overloading basically tells the compiler what must be accomplished when the increment/decrement operators are used with the object.
Also, you must pay attention to the fact that there is postfix incrementing/decrementing and prefix incrementing/decrementing which becomes very important with the notion of iterators and you should note that the syntax for overloading these two type of operators is different from each other. Here is how you can overload these operators: Overloading the increment and decrement operators
The accepted answer by Michael Burr is quite good in explaining the technique, but from the comments it seems that besides the 'how' you are interested in the 'why'. The main reasons to provide operator overloads for a given type are improving readability and providing a required interface.
If you have a type for which there is a single commonly understood meaning for an operator in the domain of your problem, then providing that as an operator overload makes code more readable:
std::complex<double> a(1,2), b(3,4), c( 5, 6 );
std::complex<double> d = a + b + c; // compare to d = a.add(b).add(c);
std::complex<double> e = (a + d) + (b + c); // e = a.add(d).add( b.add(c) );
If your type has a given property that will naturally be expressed with an operator, you can overload that particular operator for your type. Consider for example, that you want to compare your objects for equality. Providing operator== (and operator!=) can give you a simple readable way of doing so. This has the advantage of fulfilling a common interface that can be used with algorithms that depend on equality:
struct type {
type( int x ) : value(x) {}
int value;
};
bool operator==( type const & lhs, type const & rhs )
{ return lhs.value == rhs.value; }
bool operator!=( type const & lhs, type const & rhs )
{ return !lhs == rhs; }
std::vector<type> getObjects(); // creates and fills a vector
int main() {
std::vector<type> objects = getObjects();
type t( 5 );
std::find( objects.begin(), objects.end(), t );
}
Note that when the find algorithm is implemented, it depends on == being defined. The implementation of find will work with primitive types as well as with any user defined type that has an equality operator defined. There is a common single interface that makes sense. Compare that with the Java version, where comparison of object types must be performed through the .equals member function, while comparing primitive types can be done with ==. By allowing you to overload the operators you can work with user defined types in the same way that you can with primitive types.
The same goes for ordering. If there is a well defined (partial) order in the domain of your class, then providing operator< is a simple way of implementing that order. Code will be readable, and your type will be usable in all situations where a partial order is required, as inside associative containers:
bool operator<( type const & lhs, type const & rhs )
{
return lhs < rhs;
}
std::map<type, int> m; // m will use the natural `operator<` order
A common pitfall when operator overloading was introduced into the language is that of the 'golden hammer' Once you have a golden hammer everything looks like a nail, and operator overloading has been abused.
It is important to note that the reason for overloading in the first place is improving readability. Readability is only improved if when a programmer looks at the code, the intentions of each operation are clear at first glance, without having to read the definitions. When you see that two complex numbers are being added like a + b you know what the code is doing. If the definition of the operator is not natural (you decide to implement it as adding only the real part of it) then code will become harder to read than if you had provided a (member) function. If the meaning of the operation is not well defined for your type the same happens:
MyVector a, b;
MyVector c = a + b;
What is c? Is it a vector where each element i is the sum of of the respective elements from a and b, or is it a vector created by concatenating the elements of a before the elements of b. To understand the code, you would need to go to the definition of the operation, and that means that overloading the operator is less readable than providing a function:
MyVector c = append( a, b );
The set of operators that can be overloaded is not restricted to the arithmetic and relational operators. You can overload operator[] to index into a type, or operator() to create a callable object that can be used as a function (these are called functors) or that will simplify usage of the class:
class vector {
public:
int operator[]( int );
};
vector v;
std::cout << v[0] << std::endl;
class matrix {
public:
int operator()( int row, int column );
// operator[] cannot be overloaded with more than 1 argument
};
matrix m;
std::cout << m( 3,4 ) << std::endl;
There are other uses of operator overloading. In particular operator, can be overloaded in really fancy ways for metaprogramming purposes, but that is probably much more complex than what you really care for now.
Another use of operator overloading, AFAIK unique to C++, is the ability to overload the assignment operator. If you have:
class CVector
{
// ...
private:
size_t capacity;
size_t length;
double* data;
};
void func()
{
CVector a, b;
// ...
a = b;
}
Then a.data and b.data will point to the same location, and if you modify a, you affect b as well. That's probably not what you want. But you can write:
CVector& CVector::operator=(const CVector& rhs)
{
delete[] data;
capacity = length = rhs.length;
data = new double[length];
memcpy(data, rhs.data, length * sizeof(double));
return (*this);
}
and get a deep copy.
Operator overloading allows you to give own meaning to the operator.
For example, consider the following code snippet:
char* str1 = "String1";
char* str2 = "String2";
char str3[20];
str3 = str1 + str2;
You can overload the "+" operator to concatenate two strings. Doesn't this look more programmer-friendly?