why the template argument deduction/substitution failed in the code?- - c++

Question

I am using compiler g++ 6.3.0 (c++14).
In the code-
#include<iostream>
int f(auto a){return a;}
int f1(auto (*g)(int),int a) {return g(a);}
main()
{
std::cout<< f1(f,8);
}
Compiler is not able to deduce the return type of g.
It shows following error-
temp.cpp: In function 'int main()':
temp.cpp:9:20: error: no matching function for call to 'f1(<unresolved overloaded function type>, int)'
std::cout<< f1(f,8);
^
temp.cpp:5:5: note: candidate: template<class auto:2> int f1(auto:2 (*)(int), int)
int f1(auto (*g)(int),int a) {return g(a);}
^~
temp.cpp:5:5: note: template argument deduction/substitution failed:
temp.cpp:9:20: note: couldn't deduce template parameter 'auto:2'
std::cout<< f1(f,8);
^
But no error comes in the code-
#include<iostream>
int f(int /* <<<<< */ a){return a;} // only (auto a) is changed to (int a)
int f1(auto (*g)(int),int a) {return g(a);}
main()
{
std::cout<< f1(f,8);
}
Help me understand the error...

Answer 1

int f(auto a){return a;}
is equivalent to
template <typename T>
int f(T a){return a;}
You cannot take the address of a template (or overload set) - that is why you're seeing that error. Workarounds:
Take the address of the instantiation you want:
return f1(f<int>,8);
Make f1 accept auto and pass a lambda:
int f1(auto g, int a) {return g(a);}
int main()
{
std::cout<< f1([](auto x){ f(x); },8);
}