Quick A: Yes, std::ref can be reassigned (see example)

Recently on SO:

Is this std::ref behaviour logical?

A small modification to f2 provides the clue:

template<class T>
void f2(T arg)
{
    arg.get() = xx;
}

This now does what you expect.

This has happened because std::ref returns a std::reference_wrapper<> object. The assignment operator of which rebinds the wrapper. (see http://en.cppreference.com/w/cpp/utility/functional/reference_wrapper/operator%3D)

It does not make an assignment to the wrapped reference.

In the f1 case, all is working as you expected because a std::reference_wrapper<T> provides a conversion operator to T&, which will bind to the implicit right hand side of ints implicit operator+.