C ++ Overload Conversion Operators

Question

I am trying to create a class that allows for implicit casting of some built-in types, such as unsigned long int, and since I am trying to make it as correct as possible (this is my first important C ++ project), I have got into a strange problem regarding correctness const:

It works:

#include <iostream> class CustomizedInt { private: int data; public: CustomizedInt(); CustomizedInt(int input); operator unsigned long int () const { unsigned long int output; output = (unsigned long int)data; return output; } }; CustomizedInt::CustomizedInt() { this->data = 0; } CustomizedInt::CustomizedInt(int input) { this->data = input; } int main() { CustomizedInt x; unsigned long int y = x; std::cout << y << std::endl; return 0; } 

But this:

 #include <iostream> class CustomizedInt { private: int data; public: CustomizedInt(); CustomizedInt(int input); operator unsigned long int () const; }; CustomizedInt::CustomizedInt() { this->data = 0; } CustomizedInt::CustomizedInt(int input) { this->data = input; } CustomizedInt::operator unsigned long() { unsigned long int output; output = (unsigned long int)data; return output; } int main() { CustomizedInt x; unsigned long int y = x; std::cout << y << std::endl; return 0; } 

gives me this error in Visual Studio 2010: error C2511: 'CustomizedInt::operator unsigned long(void)' : overloaded member function not found in 'CustomizedInt'

Now, if I remove the const keyword from the operator definition, everything is fine. This is mistake? I read that I should use the const keyword after each (public) method / statement to clearly state that it does not modify the current object in any way.

Also, I know that defining such an operator can be bad practice, but I'm not sure I fully understand the warnings associated with it. Can someone please outline them? Would it be better if you just define a public method called ToUnsignedLongInt?

Answer 1

The signature of the function does not match the definition of the function.

 operator unsigned long int () const; 

and

 CustomizedInt::operator unsigned long() { ... } ^^^ const missing 

In this case, you should mark the conversion operator as const , since it does not affect the internal state of the object.

Also, use constructor initialization lists to initialize member variables.

 CustomizedInt::CustomizedInt() : data() { } CustomizedInt::CustomizedInt(int input) : data(input) { } 

Answer 2

You can remove const from the declaration, but what you almost certainly want to do is add it to the definition:

 CustomizedInt::operator unsigned long() const { unsigned long int output; output = (unsigned long int)data; return output; } 

Answer 3

Yes, if your member function does not affect the logical state of the object, then you really need to postfree it with const in order for the compiler to provide this.

But in this case you need to add const when you define the function body!

Answer 4

You just need to copy the same function prototype into the implementation. i.e.

 CustomizedInt::operator unsigned long int() const