Functions and function pointers in C++ - c++

Question

With reference to the following code
#include <iostream>
using std::cout;
using std::endl;
#include <vector>
using std::vector;
void function() {
cout << "Hello World" << endl;
}
int main() {
vector<void (*) ()> functions;
functions.push_back(function); // (1) no error
functions.push_back(&function); // (2) no error
for (const auto& func : functions) {
func();
}
// vector<decltype(function)> vec; // (3) error
return 0;
}
There seems to be an error when I uncomment (3), I am just trying to understand the reasoning behind this. When I pass in a function as the argument to a templated function does it resolve the type to a function pointer? It would make sense for the compiler to deduce all function types as such to a function pointer but why then does the decltype() not resolve to a function pointer?

Answer 1

decltype(function) is void() - a function.
what you need is the decayed version of a function - void(*)():
std::decay<decltype(function)>::type
std::vector < std::decay<decltype(function)>::type > myPtrFunctionVec;
PS.
if you're working with VC++ (visual stdio) you can easily see the type deduced from decltype by printing typeid(decltype(XXX)).name(). VC++, unlike other compilers, gives the undecorated name of a type. very handy for metaprogramming debugging.
EDIT:
as #Daniel Jour has commented, the solution decltype(&function) workd as well, because the construct &f gives the pointer to f, which is what you need