Template function as a template argument

With generic lambda from C++14 you might do:

template<typename T> void a(T t) { /* do something */}
template<typename T> void b(T t) { /* something else */ }

template <typename F>
void function(F&& f) {
    f(someobj);
    f(someotherobj);
}

void test() {
    // For simple cases, auto&& is even probably auto or const auto&
    function([](auto&& t){ a(t); });
    function([](auto&& t){ b(t); });

    // For perfect forwarding
    function([](auto&& t){ a(std::forward<decltype(t)>(t)); });
    function([](auto&& t){ b(std::forward<decltype(t)>(t)); });
}

Can compilers still inline the calls if they are made via function pointers?

They can, but it is indeed more complicated, and they may fail more often than with functor or template.

In order to solve this problem with templates, you have to use a template template parameter. Unfortunately, you cannot pass template template function as a type, because it has to be instantiated first. But there is a workaround with dummy structures. Here is an example:

template <typename T>
struct a {

    static void foo (T = T ())
    {
    }

};

template <typename T>
struct b {

    static void foo (T = T ())
    {
    }

};

struct SomeObj {};
struct SomeOtherObj {};

template <template <typename P> class T>
void function ()
{
    T<SomeObj>::foo ();
    T<SomeOtherObj>::foo ();
}

int main ()
{
    function<a>();
    function<b>();
}