Q:

Can this member function selection code be written without std::invoke? 

I was trying to select a member fn based on some constexpr value.  I then tried to call the selected function, but I was getting errors about how I am calling member fn with incorrect syntax. 
error: must use '.*'  or '->*' to call pointer-to-member function in
'S::SelectedGetter<&S::fn1, &S::fn2>::fn (...)', e.g. '(... ->*
S::SelectedGetter<&S::fn1, &S::fn2>::fn) (...)'     
    18 |     return SelectedGetter<&S::fn1, &S::fn2>::fn();

I tried to call it "properly" but failed.  In the end I am using std::invoke, but I wonder if this can be done without std::invoke, using just "raw" C++ syntax. 
#include <algorithm>
#include <type_traits>

static constexpr int number = 18;

struct S
{
    using GetterFn = uint32_t(S::*)() const;
    uint32_t fn1()const {
        return 47;
    }
    uint32_t fn2() const {
        return 8472;
    }

    template <GetterFn Getter1, GetterFn Getter2>
    struct SelectedGetter
    {
        static constexpr GetterFn fn = (number < 11) ?  Getter1 : Getter2;
    };

    uint32_t f() {
        return std::invoke((SelectedGetter<&S::fn1, &S::fn2>::fn), this);
    }
};

int main() 
{
    return S{}.f() % 100;
}

godbolt link
Note: I am fine with C++20 solutions, for example if some concepts magic can help...

A:

You can call it like normal member function pointer call. 
The correct syntax would be
 return ((*this).*SelectedGetter<&S::fn1, &S::fn2>::fn)();

or
return (this->*SelectedGetter<&S::fn1, &S::fn2>::fn)();

(See a demo)

Side notes:

If the functions you call in f are const, also you could make it also uint32_t f() const
Secondly, you can replace the SelectedGetter with a variable template (since c++14), and now you need less-typing

It will look like
// variable template
template<GetterFn Getter1, GetterFn Getter2>
static constexpr auto fn = (number < 11) ?  Getter1 : Getter2;

uint32_t f() const {
   return (this->*fn<&S::fn1, &S::fn2>)();
}

(See a demo)

