Re: design question with using call_traits and functors

Nat Goodspeed schrieb:
> Hansi wrote:
> 
>> I want to make a functor which uses automatically an reference type if 
>> the provided value isn't a pointer.
>>
>> Now the only thing that I don't like very much is that the user has to 
>> provide the template argument for setter. Is there some possibility to 
>> solve it that the template argument is automatically deduced?
>> Important is that I have to possible calls one when the value is 
>> provided as value -> I need to have a reference parameter and the other 
>> is when a pointer is provided.
> 
> The following works for me:
> 
> #include <iostream>
> #include <boost/lexical_cast.hpp>
> 
> template<typename Value>
> void Set(const std::string& newVal, Value* pVal)
> {
>      *pVal = boost::lexical_cast<Value>(newVal);
> }
> 
> template<typename Value>
> void Set(const std::string& newVal, Value& refVal)
> {
>      refVal = boost::lexical_cast<Value>(newVal);
> }
> 
> int main(int argc, char *argv[])
> {
>      int target;
>      Set("17", &target);
>      std::cout << "Set(Value*) produced " << target << '\n';
>      Set("34", target);
>      std::cout << "Set(Value&) produced " << target << '\n';
>      return 0;
> }

This version I have tested also, the only Problem which could happen I 
think is the reference of reference problem. Isn't it?

Best regards
Hansjörg

> 
> If you want those overloaded template free functions to forward to a 
> functor, that should be fine too: as you see, inside them you can 
> explicitly name the inferred type.