What operator int() means in C++?

Short answer to the title of this post: It's a user-defined conversion to int.

Long answer: read on ;-).  The code below is a sample of the user-defined conversion. 

class MyClass {
public:
   //..
   operator HANDLE() const { return(m_hCJ); }
   //..


private: 
   HANDLE m_hCJ;          // handle to volume
   //..

};

The code above would return MyClass object's m_hCJ member value if a conversion to HANDLE type is requested. For example:

MyClass testClass;
HANDLE testHandle = testClass;

In the preceding code, testClass operator HANDLE() will be called (at runtime?) to return testClass.m_hCJ value.

For more comprehensive example and explanation, see: http://en.cppreference.com/w/cpp/language/cast_operator.

I brought this issue up because it uses the C++ operator keyword which is usually used for operator overloading. It could confuse those who hasn't seen C++ code that uses user-defined conversion.

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