Conversion Operators in C++

Question

Please help me understand how exactly the conversion operators in C++ work. I have a simple example here which I am trying to understand, though it is not very clear how the conversion actually happens by the compiler.

class Example{ public:     Example();     Example(int val);     operator unsigned int();     ~Example(){} private:     int itsVal; };  Example::Example():itsVal(0){}  Example::Example(int val):itsVal(val){}  Example::operator unsigned int (){     return (itsVal); }  int main(){     int theInt = 5;     Example exObject = theInt; // here      Example ctr(5);     int theInt1 = ctr; // here     return 0; } 

Answer 1

You can walk through that code with a debugger (and/or put a breakpoint on each of your constructors and operators) to see which of your constructors and operators is being invoked by which lines.

Because you didn't define them explicitly, the compiler also created a hidden/default copy constructor and assignment operator for your class. You can define these explicitly (as follows) if you want to use a debugger to see where/when they are being called.

Example::Example(const Example& rhs) : itsVal(rhs.itsVal) {}  Example& operator=(const Example& rhs) {     if (this != &rhs)     {         this->itsVal = rhs.itsVal;     }     return *this; }