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
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.