C ++运算符重载综合转换

Question

This example is from "Thinking in C++", I have one question regarding compiler synthesizing the operator conversion function.

Question
When object of class Four is passed (in the function call f()), the overload operation () is called. But I am not able to make out the logic used (compiler synthesizes the operation call) by compiler to achieve this conversion.

At max, I can expect explicit conversion behavior, like
1. obj3 = (Three)obj4;
2. obj3 = Three(obj4);
3. obj3 = static_cast <Three > (obj4);

Now for any one of the above conversion - how does the compiler synthesize,
(Three) obj4.operator()?

May be I am missing some major point.

Example

//: C12:Opconv.cpp
// Op overloading conversion

class Three {
  int i;
public:
  Three(int ii = 0, int = 0) : i(ii) {}
};

class Four {
  int x;
public:
  Four(int xx) : x(xx) {}
  operator Three() const { return Three(x); }
};

void g(Three) {}

int main() {
  Four four(1);
  g(four);
  g(1);  // Calls Three(1,0)
} ///:~

Answer 1

First of all it is not operator() which you have provided, it is operator Three . This operator tells the compiler how to convert an object of class Four to an object of class Three . In g(four) call compiler is using this operator since the function g is expecting an argument of type Three . Since there is an conversion available compiler is using it. In the second case, since the constructor of Three is not declared as explicit and it is possible to construct a object of class Three using a single integer (using Three constructor) compiler is using that constuctor to create an object of the class Three so that function g can be called.

Answer 2

First of all, class Four does not contain an operator() , but it does have an operator Three() , which is a conversion operator.

In the line

g(four);

the compiler needs to convert four to an object of class Three and synthesises a call to operator Three() to perform that conversion. The synthesised conversion is equivalent to

g(four.operator Three());