std::bind and function templates

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();
}