Hello, I meet a question about conversion operator overload,please
help me, thanks. Here is the program:

struct return_overload
{
  unsigned int data;
  return_overload(unsigned int init_data):data(init_data){};

  template<typename T>
  operator T() const {std::cout << "operator() T"<<std::endl;
                      return T(double(data)/2);}

  operator unsigned int () const {std::cout << "operator 
             unsigned int()" << std::endl;return data >> 1;}
};
return_overload binary(unsigned int data)
{
  return return_overload(data);
}

int main(int argc, char* argv[])
{
double a;
unsigned int b;
a = binary(123);
b = binary(123);
std::cin.get();
return 0;
}

the output under g++, CC and VC.net are:
operator unsigned int
operator unsigned int

I want to know why the compiler don't pick the template version.
thanks.
andyzhang from china.
11/2/2003
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 2 Nov 2003 07:12:06 -0500 ， zhstrongman <zhstrongman@sina.com> : 

>Hello, I meet a question about conversion operator overload,please
>help me, thanks. Here is the program:
>
>struct return_overload
>{
>  unsigned int data;
>  return_overload(unsigned int init_data):data(init_data){};
>
>  template<typename T>
>  operator T() const {std::cout << "operator() T"<<std::endl;
>                      return T(double(data)/2);}
>
>  operator unsigned int () const {std::cout << "operator 
>             unsigned int()" << std::endl;return data >> 1;}
>};
>return_overload binary(unsigned int data)
>{
>  return return_overload(data);
>}
>
>int main(int argc, char* argv[])
>{
>double a;
>unsigned int b;
>a = binary(123);
>b = binary(123);
>std::cin.get();
>return 0;
>}
>
>the output under g++, CC and VC.net are:
>operator unsigned int
>operator unsigned int
>
>I want to know why the compiler don't pick the template version.
>thanks.
this is not compiler's bug.
the compiler will not maching a template conversion operator T() with
this program, template will not conclude which conversion what you
needed with your return value.
you can wrote conversion explicit, for ex:

return_overload ra(10);
ra.template operator double<double>();

i am chinese programmer too, visit http://coneos.126.com  :-)


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zhstrongman wrote:
> Hello, I meet a question about conversion operator overload,please
> help me, thanks. Here is the program:
> 
> struct return_overload
> {
>   unsigned int data;
>   return_overload(unsigned int init_data):data(init_data){};
> 
>   template<typename T>
>   operator T() const {std::cout << "operator() T"<<std::endl;
>                       return T(double(data)/2);}
> 
>   operator unsigned int () const {std::cout << "operator 
>              unsigned int()" << std::endl;return data >> 1;}
> };
> return_overload binary(unsigned int data)
> {
>   return return_overload(data);
> }
> 
> int main(int argc, char* argv[])
> {
> double a;
> unsigned int b;
> a = binary(123);
> b = binary(123);
> std::cin.get();
> return 0;
> }
> 
> the output under g++, CC and VC.net are:
> operator unsigned int
> operator unsigned int
> 
> I want to know why the compiler don't pick the template version.

It works the way you expect in VC++ 7.1 (Visual Studio .NET 2003).
I think the other compilers you tried are incorrect.

Making operator unsigned int an out-of-line specialisation of
template operator T works in both g++ 3.0 and VC++ 7.1, but not
in earlier versions of VC++:

    template<>
    return_overload::operator unsigned int () const {
        std::cout << "operator unsigned int()" << std::endl;
        return data >> 1;
    }

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