[Boost-users] [type_traits] deduce argument type of member function with function_traits

Subject: [Boost-users] [type_traits] deduce argument type of member function with function_traits
From: alfC (alfredo.correa_at_[hidden])
Date: 2011-09-03 01:29:57

Is function_traits the right way to extract the argument type of a member 
function? 

In the following example I want to make the last line work. This will be 
part of larger question, on how to deduce the argument type of operator() 
for function objects in which operator() is not overloaded.

#include <typeinfo>
#include <iostream>
#include <boost/type_traits/function_traits.hpp>
#include <boost/type_traits/remove_pointer.hpp>

using namespace std;

double dsquare(double l){ return l*l; }
int isquare(int l){ return l*l; }

struct csquare_impl{
    int do(char c) const{
        return 'c';
    }
};
csquare_impl csquare;

template<class F>
void test(F f){
    typename boost::function_traits<typename 
boost::remove_pointer<F>::type>::arg1_type test_var;
    cout << typeid(test_var).name() << endl;
}

int main(){
    test(dsquare); // ok, prints "d"(double)
    test(isquare); // ok, prints "i"(int)
    test(csquare_impl::do);  // not ok, want to print "c"(char)
    return 0;
}

Thanks,
Alfredo