design question with using call_traits and functors
Hello, I want to make a functor which uses automatically an reference type if the provided value isn't a pointer. Now I have thougth to use the call_traits library. template<typename Value> struct Setter { typedef void result_type; void operator()(const std::wstring& newVal, Value* pVal) { *pVal = boost::lexical_cast<typename Value>(newVal); } void operator()(const std::wstring& newVal, typename boost::call_traits<Value>::reference refVal) { refVal = boost::lexical_cast<typename Value>(newVal); } }; To call the function I have the following statement: SetValue setter = boost::bind(Setter<short>(),_1,x); 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. It would be great if someone can give me a hint. Best regards Hansjörg
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; } 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.
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.
Hansi wrote:
This version I have tested also, the only Problem which could happen I think is the reference of reference problem. Isn't it?
Um - I don't understand the bad case you're imagining. I added this: void func(int& param) { Set("51", param); } and these two lines: func(target); std::cout << "Set(ref) produced " << target << '\n'; and got what I expected. So I'm not yet following your train of thought.
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Nat Goodspeed