[type_traits] deduce argument type of member function with function_traits
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
May be: http://www.boost.org/doc/libs/1_47_0/libs/function_types/doc/html/boost_func... ? 2011/9/3 alfC <alfredo.correa@gmail.com>
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
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On Saturday, September 3, 2011 1:41:41 AM UTC-7, niXman wrote:
May be: http://www.boost.org/doc/libs/1_47_0/libs/function_types/doc/html/boost_func...
?
good to know! but I still get an error, on this new version. error: no matching function for call to ‘test(<unresolved overloaded function type>)’ /tmp/tt.cpp:25:6: note: candidate is: void test(F) [with F = char (csquare_impl::*)(char)const] but "run" is not overloaded !, and I don't know how to cast to resolve the overload. #include <typeinfo> #include <iostream> #include <boost/type_traits/function_traits.hpp> #include <boost/type_traits/remove_pointer.hpp> #include <boost/function_types/parameter_types.hpp> #include <boost/mpl/vector.hpp> #include <boost/mpl/int.hpp> #include <boost/mpl/at.hpp> using namespace std; double dsquare(double l){ return l*l; } int isquare(int l){ return l*l; } struct csquare_impl{ char run(char c) const{return 'c';} }; csquare_impl csquare; template<class F> void test(F f){ // works // typename boost::function_traits<typename boost::remove_pointer<F>::type>::arg1_type test_var; // works too, more elegant(?) typename boost::mpl::at_c<typename boost::function_types::parameter_types<F>::type, 0 >::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.csquare_impl::run); // not ok, want to print "c"(char) return 0; } BTW, this type traits is a mess, everything it is all over the place, Boost.FunctionTypes, Boost.TypeTraits.FunctionTraits, Boost.Functional.
2011/9/3 alfC <alfredo...@gmail.com>
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
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http://liveworkspace.org/code/86e774c3b3de9d7173bb16c00d56eddb 2011/9/3 alfC <alfredo.correa@gmail.com>
On Saturday, September 3, 2011 1:41:41 AM UTC-7, niXman wrote:
May be: http://www.boost.org/doc/**libs/1_47_0/libs/function_** types/doc/html/boost_**functiontypes/reference/** decomposition.html#boost_**functiontypes.reference.** decomposition.parameter_types<http://www.boost.org/doc/libs/1_47_0/libs/function_types/doc/html/boost_functiontypes/reference/decomposition.html#boost_functiontypes.reference.decomposition.parameter_types>
?
good to know! but I still get an error, on this new version.
error: no matching function for call to ‘test(<unresolved overloaded function type>)’ /tmp/tt.cpp:25:6: note: candidate is: void test(F) [with F = char (csquare_impl::*)(char)const]
but "run" is not overloaded !, and I don't know how to cast to resolve the overload.
#include <typeinfo> #include <iostream> #include <boost/type_traits/function_traits.hpp> #include <boost/type_traits/remove_pointer.hpp> #include <boost/function_types/parameter_types.hpp> #include <boost/mpl/vector.hpp> #include <boost/mpl/int.hpp> #include <boost/mpl/at.hpp>
using namespace std;
double dsquare(double l){ return l*l; } int isquare(int l){ return l*l; }
struct csquare_impl{ char run(char c) const{return 'c';}
}; csquare_impl csquare;
template<class F> void test(F f){ // works // typename boost::function_traits<typename boost::remove_pointer<F>::type>::arg1_type test_var; // works too, more elegant(?) typename boost::mpl::at_c<typename boost::function_types::parameter_types<F>::type, 0 >::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.csquare_impl::run); // not ok, want to print "c"(char) return 0; }
BTW, this type traits is a mess, everything it is all over the place, Boost.FunctionTypes, Boost.TypeTraits.FunctionTraits, Boost.Functional.
2011/9/3 alfC <alfredo...@gmail.com>
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
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