c++ - std::function in template function -

i trying write template function can accept functor parameter , call afterwards. program follows:

#include <iostream> #include <functional> using namespace std;  template<typename r, typename... args> r call(function<r(args...)> fun, args... args) {     cout << "call@ " << __line__ <<endl;     return fun(args...); }  int main() {     cout << call(std::plus<int>(),1,2) <<endl;     return 0; } 

the g++ compplains:

g++ -c -wall -std=c++0x -i../include a.cpp -o a.o a.cpp: in function ‘int main()’: a.cpp:16:38: error: no matching function call ‘call(std::plus<int>, int, int)’ a.cpp:16:38: note: candidate is: a.cpp:7:3: note: template<class r, class ... args> r call(std::function<_res(_argtypes ...)>, args ...) a.cpp:7:3: note:   template argument deduction/substitution failed: a.cpp:16:38: note:   ‘std::plus<int>’ not derived ‘std::function<_res(_argtypes ...)>’ make: *** [a.o] error 1 

i suppose std::plus<int>() deduced std::function<int(int,int)>, didn't. why that? gcc gcc version 4.7.2 20120921 (red hat 4.7.2-2) (gcc)

i suppose std::plus() deduced std::function

no. not deduced given have passed object of type std::plus<int>.

in case, you not need use std::function, use when storing different functions/function objects can called specific signature.

with that, can have call function accept function/function object directly, original type deduced, without using std::function. also, might want use perfect forwarding when accepting parameters , use std::forward when passing them arguments function/function object. should use return type of function return type of call. use c++11's trailing return type decltype that.

#include <iostream> #include <functional> using namespace std;  template<typename r, typename... args> auto call(r fun, args&&... args) -> decltype(fun(std::forward<args>(args)...)) {     cout << "call@ " << __line__ <<endl;     return fun(std::forward<args>(args)...); }  int main() {     cout << call(std::plus<int>(),1,2) <<endl;     return 0; } 

live code

---

as @jan hudec has commented, __line__ in there result same in calls call, whatever function passed.