Question

// some arbitrary function 
template<typename T>
void log( T&& obj )
{
    std::cout << obj << std::endl;    
}

// arbitrary transformation
template<typename T>
T convert(T&& obj) { 
     return obj; 
}

template<template <typename> typename F, typename... T>
void callOn( F<T> func,  ///  issue: what's the type of func?                 
             T&&... params)
{
    func(std::forward<T>(convert(std::forward<T>(params)))...);
}

int main()
{   
    callOn(log, -1, -2.0);      
     return 0;
}

Is this possible at all?

Compiler complains: no matching function for call to ‘callOn(, ..>). Why ?

Update: suppose log is not unary function

template<typename T>
void log(T&& value) { std::cout << value << std::endl; }

template<typename First, typename... Rest>
void log(First&& f, Rest&&... rest)
{
    std::cout << f << ",";
    log(std::forward<Rest>(rest)...);
}

callOn takes type “template ” which doesn’t match the type of log? How to specify the type of func?

Answer 1

Use a function object. Here’s a compilable example:

#include <utility>
#include <iostream>

struct Log
{
    template<typename T> void operator()(T&& t) { 
        std::cout << t << std::endl; 
    }

    template<typename T, typename... Rest> void operator()(T&& t, Rest&&... rest)
    {
        std::cout << t << ", ";
        (*this)(std::forward<Rest>(rest)...);
    }
};

template<typename T>
T convert(T&& obj) {
     return obj;
}

template<typename F, typename... T>
void callOn(F funcobj, T&&... params)
{
    funcobj(std::forward<T>(convert(std::forward<T>(params)))...);
}

int main()
{
    callOn(Log(), -1, -2.17, "abc");
    return 0;
}