Effective c++ 3/e item 15 疑問解惑

轉載自http://stackoverflow.com/questions/4728160/thrown-off-by-functor-syntax-in-effective-c

The first

operator FontHandle() const 
    {return f;}

The second

FontHandle operator()() const
    {return f;}

The first, operator FontHandle, is a conversion operator. It allows an instance of this class type to be implicitly converted to a FontHandle object, so you could write:

Font myFont; FontHandle handle = myFont;

More commonly, conversion operators are used to allow you to use an object of one type as if it were another type in an expression. For example,

void f(FontHandle fh); Font myFont; f(myFont); // converts myFont to a FontHandle via the conversion operator

The second, operator(), is an overload of the function call operator. It allows an instance of your class type to be used as if it were a function taking no arguments:

Font myFont; myFont();