I'm trying to overload the comma operator with a non-friend non-member function like this:
#include <iostream>
using std::cout;
using std::endl;
class comma_op
{
int val;
public:
void operator,(const float &rhs)
{
cout << this->val << ", " << rhs << endl;
}
};
void operator,(const float &lhs, const comma_op &rhs)
{
cout << "Reached!\n"; // this gets printed though
rhs, lhs; // reversing this leads to a infinite recursion ;)
}
int main()
{
comma_op obj;
12.5f, obj;
return 0;
}
Basically, I'm trying to get the comma operator usable from both sides, with a float. Having a member function only allows me to write obj, float_val
, while having an additional helper non-friend non-member function allows me to write float_val, obj
; but the member operator function doesn't get called.
GCC cries:
comma.cpp: In function ‘void operator,(const float&, const comma_op&)’:
comma.cpp:19: warning: left-hand operand of comma has no effect
comma.cpp:19: warning: right-hand operand of comma has no effect
Note:
I realise that overloading operators, that too to overload comma op., Is confusing and is not at all advisable from a purist's viewpoint. I'm just learning C++ nuances here.
In C++, we can overload the comma operator using Operator Overloading. For Example: For “Send the query X to the server Y and put the result in variable Z”, the “and” plays the role of the comma. The comma operator (, ) is used to isolate two or more expressions that are included where only one expression is expected.
Overloading of function-call operator in C++ The function call operator is denoted by “()” which is used to call function and pass parameters. It is overloaded by the instance of the class known as a function object.
These operators cannot be overloaded because if we overload them it will make serious programming issues. For an example the sizeof operator returns the size of the object or datatype as an operand. This is evaluated by the compiler. It cannot be evaluated during runtime.
You cannot overload the following operators: . You cannot overload the preprocessor symbols # and ## . An operator function can be either a nonstatic member function, or a nonmember function with at least one parameter that has class, reference to class, enumeration, or reference to enumeration type.
void operator,(const float &rhs)
You need a const
here.
void operator,(const float &rhs) const {
cout << this->val << ", " << rhs << endl;
}
The reason is because
rhs, lhs
will call
rhs.operator,(lhs)
Since rhs
is a const comma_op&
, the method must be a const
method. But you only provide a non-const
operator,
, so the default definition will be used.
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