std::bind and function templates - c++

Question

I am currently trying to use std::bind to create a std::function<void()> from the function template
template<class Iterator>
void printRange(Iterator first, Iterator last) {
std::copy(first, last, std::ostream_iterator<typename Iterator::value_type>(std::cout, " "));
std::cout << std::endl;
}
Conceptually, what I want is
int main() {
std::vector<int> v{1, 2, 3};
auto f0 = std::bind(printRange, v.begin(), v.end()); // won't compile, of course
f0();
return 0;
}
I understand that this does not compile and I have to instantiate the function template before I can actually use it. For example, the following alternatives would work:
auto f1 = std::bind(printRange<std::vector<int>::const_iterator>, v.begin(), v.end());
auto f2 = std::bind(printRange<decltype(v.begin())>, v.begin(), v.end());
auto f3 = [&v]() { printRange(v.begin(), v.end()); };
I already created a convenience function
template<class Iterator>
std::function<void()> makePrintRangeFunction(Iterator first, Iterator last) {
return std::bind(printRange<Iterator>, first, last);
}
to ease the process:
auto f4 = makePrintRangeFunction(v.begin(), v.end());
I wonder if it is possible to create a more generic std::function<void()> generator, accepting a function template as a first argument and the function template arguments as a variable-length argument list? If not using built-in language features, maybe via a macro?

Answer 1

As long as you do not need to have template function return type, you can do this:
#include <functional>
#include <iostream>
#include <typeinfo>
template<typename ... T>
std::function<void()> makePrintRangeFunction(void (*f)(T...), T... param) {
return std::bind(f, param...);
}
template<typename T, typename V>
void print(T type, V val)
{
std::cout << typeid(type).name() << '\n' << val << '\n';
}
int main()
{
int i = 5;
double d = 10.5;
auto f = makePrintRangeFunction(print, i, d);
f();
}

Answer 2

Maybe the following code will help :)
template <class F, class... Args>
void test(F&& f, Args&&... args) {
std::function<typename std::result_of<F(Args...)>::type()> task(
std::bind(std::forward<F>(f), std::forward<Args>(args)...));
task();
}

Answer 3

If your compiler supports C++14 you could define a generic lambda wrapper as:
template<typename F>
auto fungen(F f) {
return [=](auto... args) { f(args...); };
}
Use case:
int main() {
std::vector<int> v {1, 2, 3, 4};
auto f = fungen(printRange<std::vector<int>::iterator>);
f(v.begin(), v.end());
}
Live Demo