Template substitution failure with std::function - c++

Question

I am trying to pass a callback function as function parameter. But getting template substitution failure errors in following code. Not sure why template substitution is failing.
#include<iostream>
#include <map>
#include <tuple>
#include <functional>
template<typename A,typename B>
void myfun(std::map<A,B> & mm, std::function<std::tuple<A,B>(void)> fn)
{
A key;
B val;
std::tie(key,val) = fn();
mm[key] = val;
}
std::tuple<std::string,int> fun()
{
return std::make_tuple(std::string("hi"),1);
}
int main()
{
std::map<std::string,int> gg;
#if 0
//fixed version
std::function<std::tuple<std::string,int>(void)> yy = fun;//fixed
myfun(gg,yy);//fixed
#else
// error causing code
myfun(gg,fun);
#endif
}
And error is as following
main.cpp:8:6: note: template argument deduction/substitution failed:
main.cpp:25:17: note: mismatched types 'std::function<std::tuple<_T1, _T2>()>' and 'std::tuple<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >, int> (*)()'
myfun(gg,fun);

Answer 1

The compiler can't both cast to a std::function and deduce the template arguments. It doesn't understand the mapping between an arbitrary function pointer and a std::function.
There are a few ways round this.
You could explicitly create a std::function at the call site:
myfun(gg,std::function<std::tuple<std::string,int>(void)>{fun});`
You could write a make_function function to deduce the types for you. You can find discussions and implementations of this online, such as here, here and here.
myfun(gg,make_function(fun));
You could just forget about std::function and deduce the entire function type. This is the approach I would take:
template<typename A,typename B, typename Fun>
void myfun(std::map<A,B> & mm, Fun fn)
{
A key;
B val;
std::tie(key,val) = fn();
mm[key] = val;
}