I'm fairly familiar with operator overloading, however I am wondering how do we implement something like this:
myClass myclassobj;
int x;
x = 5;
x = x + myclassobj
There is no way to overload the + operator for the int class, so something should be done from myClass, but how would we do that? I am probably using the wrong keywords, but searching through SO didn't lead to anithing. Apologies if I did something wrong, this is my first post here.
Edit - My class is a custom vector class, so simply converting it to given type won't work.
Define an overloaded operator with the signature int operator+(int, myClass).
You are right.
There is no way to overload the + operator for the int class, so
something should be done from myClass
Your question:
but how would we do that?
My answer:
You should use user defined type conversion. It may work with a conversion operator.
#include <iostream>
class myClass
{
int i;
public:
myClass(int i=0) : i(i) { }
operator int(){ // A conversion from myClass to int may solve your problem.
return i;
}
};
int main()
{
myClass myclassobj(99);
int x=7;
x = 5;
x = x + myclassobj;
std::cout<<x<<std::endl;
return 0;
}
Brian gave a good answer also, but it works only if the overloaded operator does not need protected or private members from the second argument or if overloaded operator declared as friend of myClass.
Related
Im trying to figure how such an operator works:
I found nothing about it searching the web...
Is it used as a casting between types?
Why do I need the = operator (in line a = b) instead of a b ?
What other uses does it have?
thanks
class A{
int a;
};
class B{
int b;
operator A() { return A(); }
};
int main(){
A a;
B b;
a = b;
return 0;
}
What you are looking at is a user defined conversion operator. IT has plenty of uses, for example. Consider a smart pointer:
class SmartPointer {
// Constructor destructor, operator* and & etc ..
}
If it were a raw pointer, we could check for it being nullptr like this:
if (ptr) {
ptr->do_something();
}
So how can we achieve the same with a smart pointer? We can define operator bool.
Another example could be something like a units class:
class Meters {
...
}
What if we want to be able to achieve this:
void some_operation(double meters);
Meters m{10};
some_operation(m);
Well we can define a conversion operator:
Meters::operator double() {
return _meters;
}
Remember when you are looking into this to check if you need the explicit specifier, it is likely that you will want to use this for most conversions.
Given the following code:
template <typename T>
struct Wrapper {
T value;
Wrapper(T val) : value(val) {}
}
int main() {
Wrapper<int> x(42);
int y = x; // need solution for y = x.value
return 0;
}
Is there a way to implement the assignment
int y = x;
so that it means y = x.value .
I know that overloading the assignment operator itself is not possible because it applies to the object on the left side of the assignment and friend function with two arguments is not allowed by the standard.
If this is not possible by overloading any other operator, or by using some special tricks, how would you implement this, except by invoking the get method provided by the Wrapper class such as:
int y = x.get();
Why not just provide an implicit conversion to T
operator T() { return value; }
This will cause the assignment to function because the compiler will attempt to convert the right side of the assignment to T. The implicit conversion will allow that to succeed
Note that this will cause other conversions to work besides assignment. For example it will now be possible to pass Wrapper<T> instances as T parameters. That may or may not work for your particular scenario
How would I overload the = operator in a way that I could execute
int someInt;
MyClass instanceOfMyClass;
someInt = instanceOfMyClass;
Where MyClass contains an integer named number?
You can't overload the operator= for the type int. What you are really looking for is the conversion operator operator int() for your MyClass. In your case, considering x to be the private member of your class:
operator int() { return x; }
I would avoid doing this in most cases as it can be difficult to tell what exactly is happening. That said:
class MyClass {
public:
operator int() { return number; }
private:
int number;
};
This creates an (implicit) conversion operator for your class.
The title basically says it all. I mainly want to do this so that I can create an object (say, a custom string object) that can initialize the parameters of other functions in other APIs. Here's an example of me trying to get a custom integer class to work:
#include <iostream>
using namespace std;
class test
{
public:
int member;
test(int i) : member(i) {}
friend int &operator=(int &i, test t);
};
int &operator=(int &i, test t)
{
return (i = t.member);
}
int main()
{
int i;
test t = 90;
cout << (i = t);
return 0;
}
Unfortunately I receive an error saying that operator= needs to be a member function. I understand the C++ standard's goal in preventing static and non-member overloads for the assignment operator from being implemented, but is there any other way to do this? Thanks for any help/suggestions!
This is not done with an assignment operator but with an overloaded typecast. This would make your main function work like expected:
#include <iostream>
using namespace std;
class test
{
public:
int member;
test(int i) : member(i) {}
operator int() const {return member;}
};
int main()
{
int i;
test t = 90;
cout << (i = t);
return 0;
}
What you are trying to do needs an conversion operator
operator int()
{
return this->member;
}
For the class you are trying to write(containing only integer members), You do not need to overload the = operator.
= operator is one of the member functions that is generated by the compiler by default for every class. Caveat is, it does a simple bit by bit copy(shallow copy) of class members, since you have only integers it should be good enough for you.
You would need to overload the = operator if you had dynamically allocated pointers as member functions, because in that case a shallow copy of those pointers would result in all the objects containing a member pointer pointing to the same dynamic memory location & if one of the object finishes it lifetime, other objects are left with a dangling pointer.
As #Tony, aptly points in out comments Shallow copy is usually bad but not always. See his comments for a scenario.
If at all you want to overload the assignment operator check out the Copy and Swap Idiom to do it right way.
You should also check out the Rule of Three.
Try this:
class test
{
public:
int member;
test(int i) : member(i) {}
operator int() {return this->member;}
};
int main(void)
{
int i;
test t = 90;
cout << (i = t);
return 0;
}
The assignment operator cannot be a friend function. The assignment operator can only be declared as a non-static member function. This is to ensure that it receives the L-value as its first operand. The same is true for the [], (), and -> operators. In your case, since int is an build-in type, you cannot use member function. You can implement operator int() to cast your user-defined type to int.
The assignment operator can be declared as
T& operator= (const t&);
in a class, but the arithmetic operators cannot be defined that way. It has to be friend function. I don't understand why? Can you please explain ?
It is not mandatory that arithmetic operators should be friend
Well you can define like this:
MyClass MyClass::operator + (const MyClass& t) const
{
MyClass ret(*this);
ret += t;
return ret;
}
The a + b is really a syntax sugar, the compiler will expand it to a.operator+(b). The previous sample will work if all your objects are MyClass instances, but will not work if you have to operate with others types, ie 1 + a, will not work, this can be solved by using friends.
MyClass operator + (int i, const MyClass& t)
{
MyClass ret(i);
ret += t;
return ret;
}
This has to be done when the left hand side of the + operator is not a class, or it is a class but you can't add operator + to its definition.
I think that C++ FAQ Lite will give you a definitive answer.
They ideally should be globals and not necessarily friends, so that you can write:
yourtype v = 1;
yourtype w = 1 + v;
Since, 1 is not an object of yourtype, if the operator+ were a member it would throw a fit. However, making it global makes it convert the 1 to yourtype and then perform the operation. Making it a friend helps to extract and manipulate the members of yourtype as required -- though not required. As an example: You can implement the member function operator+= and use it in the implementation of the operator+.
The problem is that if you do something like this:
class A
{
A& operator+(int n);
// ...
}
int main()
{
A my_var;
int integer = 1;
A + integer; // compiles
integer + A // error: no function operator+(int, A) defined
}
it will not compile.
A solution is to define operator+(int, A) and operator+(A, int) as friends of A.
As a side note, the Boost Operators library makes this process very easy.