Re: [boost] Preprocessor directive for a sequence that differs only in datatype

Subject: Re: [boost] Preprocessor directive for a sequence that differs only in datatype
From: Larry Evans (cppljevans_at_[hidden])
Date: 2017-07-16 15:32:31

On 07/11/2017 06:17 AM, Larry Evans via Boost wrote:
> On 07/11/2017 03:09 AM, Rishabh Arora via Boost wrote:
>> Thank you, Gavin Lambert.
>> I don't think any of the preprocessor directives will work then? because
>> type cannot be determined at comile time? Is there any other way to do
>> that?
>>
> What about replacing the switch with an index into a vector
> of function pointers?  Something like:
> 
> #include <iostream>
> template < class T >
> void print_column(std::size_t col) {
>    T dummy;
>    std::cout <<dummy<< std::endl;
>    return;
> }
> using fptr=void (*)(std::size_t);
> fptr printers[]=
> { print_column<int>
> , print_column<float>
> };
> int main()
> {
>    unsigned ncol_=2;
>    for(unsigned i=0; i<ncol_; ++i)
>    {
>      printers[i](0);
>    }
>    return 0;
> }
> 

More specifically:

#ifdef USE_PRINTER_TYPES
   template<typename... Types>
   struct printers
   {
     using fptr = void(data_frame::*)(size_t);
     static constexpr fptr _[sizeof...(Types)]=
     { &data_frame::print_column<Types>...
     };
   };
   using printer_types=printers<COLUMN_DATA_TYPES>;
#endif//USE_PRINTER_TYPES

Then, use printer_types in the data_frame::print function as:

   std::cout << "[" << column_headers_(i) << "]" << ": ";
   auto const itype=base_::operator[](column_headers_(i)).type();
#ifdef USE_PRINTER_TYPES
   printer_types::fptr printer=printer_types::_[itype];
   (this->*printer)(i);
#else
   switch(itype) {
.
.
.
#endif//USE_PRINTER_TYPES

HTH.

-regards,
Larry