Functions and function pointers in 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