Delayed function template instantiation

Question

In this rather contrived example, I'm trying to pass a function template to my function, and want my function to instantiate the function template internally. In essence, I don't want the user to know the types, and workings of my function, but just be able to pass a function template for me to instantiate myself. Self contained example (that doesn't compile):

#include <iostream>
#include <vector>
#include <iterator>
#include <algorithm>

template <template <typename InputIt, typename OutputIt> class CopyFunc>
void transferWith( CopyFunc<std::vector<int>::const_iterator,
                            std::back_insert_iterator<std::vector<int>>> const& func )
{
    std::vector<int> source{1, 2, 3, 4, 5, 6, 7};
    std::vector<int> sink;

    func(source.begin(), source.end(), std::back_inserter(sink));

    for (auto&& e : sink)
    {
        std::cout << e << std::endl;
    }
}

int main(int argc, char const *argv[])
{

    // I want to pass just the function template in,
    // and instantiate it internally
    transferWith(std::copy);

    return 0;
}

This fails to compile on gcc-4.7.2 as expected, with the following errors:

main.cpp|25 col 27 error| no matching function for call to ‘transferWith(<unresolved overloaded function type>)’
main.cpp|8 col 6 error| note: template<template<class InputIt, class OutputIt> class CopyFunc> void transferWith(const CopyFunc<__gnu_cxx::__normal_iterator<const int*, std::vector<int> >, std::back_insert_iterator<std::vector<int> > >&)
main.cpp|25 col 27 error| note:   couldn't deduce template parameter ‘template<class InputIt, class OutputIt> class CopyFunc’

Are there any tricks or indirections one can pull to get around this?

Thank you.

score 2 · Answer 1 · edited May 23 '17 at 11:43

2

Are there any tricks or indirections one can pull to get around this?

Yes. Pass a functor object with template operator()().

The code will look exactly like the version in Bart's answer.

edited May 23 '17 at 11:43

Community

1
1

answered Jan 02 '13 at 19:24

Ben Voigt

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Interesting. So it seems like you can't pass an uninstantiated function template as a template template parameter. You can only do that with class templates, hence why we need to make it a function object. Here's the code that works: http://liveworkspace.org/code/2ww7xE$8 – Alexander Kondratskiy Jan 02 '13 at 19:54

score 2 · Accepted Answer · answered Jan 02 '13 at 19:53

Just for completeness, here's an example of the functor approach:

#include <algorithm>
#include <vector>
#include <iostream>

struct Copy {
  template <typename InputIter, typename OutputIter>
  OutputIter operator()(InputIter i1, InputIter i2, OutputIter o1) {
    return std::copy(i1, i2, o1);
  }
};

template <typename CopyFunctor> 
void foo(CopyFunctor cf) {
  std::vector<int> v1 = {3, 1, 4}; 
  std::vector<int> v2(v1.size());

  cf(v1.begin(), v1.end(), v2.begin());

  for (auto i : v2) {
    std::cout << i << ' ';
  }
}

int main() {
  foo(Copy());  // displays "3 1 4"
}

Tomek · Answer 3 · 2013-01-02T19:04:04.910

Would something along the lines

template <template <typename InputIt, typename OutputIt> class CopyFunc> void transferWith()
{
    std::vector<int> source{1, 2, 3, 4, 5, 6, 7};
    std::vector<int> sink;

    CopyFunc<std::vector<int>::const_iterator, std::vector<int>::iterator>(source.begin(), source.end(), std::back_inserter(sink));

    for (auto&& e : sink)
    {
        std::cout << e << std::endl;
    }
}

called as:

transferWith<std::copy>();

do?

I am not promising this will compile out of a box...

Delayed function template instantiation

3 Answers3