operator keyword and operator overloading simulation - c++

Question

I understand that we can apply increment and decrement on intrinsic data types such as:
int a;
a++;
a--;
...etc.
However, in the following codes, if I omit the & in the line operator int& ( ) { return value; }, I will get a compile error. Please explain how and why using & makes the increment possible here:
#include <iostream>
class Foo {
public:
int value;
operator int& ( ) { return value; }
};
int main ( ) {
Foo p;
p.value = 10;
p++;
std::cout << p << std::endl;
std::cin.get ( );
return 0;
}

Answer 1

This function definition:
operator int&() { return value; }
Allows a conversion from your class Foo to a reference to int.
This line:
p++;
Cannot increment p as type Foo as the class does not have operator++ implemented. However it is able to convert p to a reference to int via a call to operator int&() which returns a reference to value within the particular instance of Foo. The reference to value is of type int which can be incremented using ++. So the code then increments value in-place.
As to the distinction between returning an in and a reference to int, the following code shows the behaviour you mention is not distinct to the conversion oeprator:
class Foo
{
public:
Foo() : m_value(0)
{
}
int getCopy() { return m_value; }
int& getRef() { return m_value; }
private:
int m_value;
};
int main()
{
Foo f;
// Works
f.getRef()++;
// Does not work on g++ with 'error: lvalue required as increment operand`
f.getCopy()++;
}
The reason you cannot do an increment when the conversion operator returns int rather than int& is that the increment operation requires an lvalue, i.e. something that can be the target of an assignment. See here for more detail.