C / C++ / MFC

George_George wrote:

Is it C++ Spec regulated? Or vendor implementation specific?

George_George wrote:

About the link, it only deals with overload, not covering template, right?

// VanillaConsole.cpp : Defines the entry point for the console application.
//
 
#include "stdafx.h"
#include <iostream>
using namespace std;
 
struct Magic
{
	operator int() const
	{
		return 1;
	}
 
	operator double() const
	{
		return 1.234;
	}
};
 
int main()
{
	Magic m;
 
	int i = m;
	double d = m;
 
	cout << "i=" << i << ", d=" << d << endl;
 
	return 0;
}

Rajkumar R wrote:

yes steve, you are correct.

But, My understanding is "operator double()" and "operator int()" are not overloaded members instead it is two different overloaded operators for user-defined conversion of prototype "double operator double()" and "int operator int()" since conversion has the same return type of the conversion type the return type of the conversion operators are redundant and compiler gives error message "C2549: user-defined conversion cannot specify a return type" actually these functions differ not by return type but by the prototype itself.

Stephen Hewitt wrote:

I'd consider them member functions since they have access to the object's state through an implicit this pointer

Stephen Hewitt wrote:

and the choice of which function is called depends on the type to which the result of the call is assigned

Rajkumar R wrote:

both are different functions and its not matched by the return type but as a typecast operator.

// VanillaConsole.cpp : Defines the entry point for the console application.
//
 
#include "stdafx.h"
#include <iostream>
using namespace std;
 
struct Magic
{
	operator int() const
	{
		cout << "int" << endl;
		return 0;
	}
 
	operator float() const
	{
		cout << "float" << endl;
		return 0.0;
	}
 
	operator double() const
	{
		cout << "double" << endl;
		return 0.0;
	}
};
 
int main()
{
	Magic m;
 
	int i = m;
	float f = m;
	double d = m;
 
	return 0;
}

Stephen Hewitt wrote:

which function is called depends on the type the expression is assigned to.

Stephen Hewitt wrote:

You can use this with templates too

// VanillaConsole.cpp : Defines the entry point for the console application.
//
 
#include "stdafx.h"
#include <iostream>
using namespace std;
 
struct Magic
{
	template <typename T>
	operator T() const
	{
		return 0.1234012340123401234;
	}
};
 
int main()
{
	Magic m;
 
	int i = m;
	float f = m;
	double d = m;
 
	cout << "i=" << i << ", f=" << f << ", d=" << d << endl;
 
	return 0;
}

// VanillaConsole.cpp : Defines the entry point for the console application.
//
 
#include "stdafx.h"
#include <iostream>
#include <typeinfo>
using namespace std;
 
struct Magic
{
	template <typename T>
	operator T() const
	{
		cout << typeid(T).name() << endl;
		return T();
	}
};
 
int main()
{
	Magic m;
 
	int i = m;
	float f = m;
	double d = m;
 
	return 0;
}

George_George wrote:

I am confused if T is int, what is T()?