C++ move semantics behaving differently based on the function return type

I have two functions taking in an rvalue reference to a string and adding something to that string. The first function returns the modified string and the second one has return type void.

Calling the first function on a string results in the original string being emptied, which is what I expected. However, calling the second function on a string brings it to its modified state as if the function were taking an lvalue reference.

Can someone please explain what is going on here? Is the compiler (VC++ configured for ISO C++ 14) doing some sort of "optimisation"?

#include <iostream>
#include <string>

using namespace std;

string g(string&& s)
{
    s += " post g call";
    return s;
}

void f(string&& s)
{
    s += " post f call";
    
}

int main()
{
    string s1("starting string");
    string s2("starting string");

    auto s1a = g(move(s1));
    f(move(s2));

    cout << s1 << endl;
    cout << s1a << endl;
    cout << s2 << endl;

}

s1 is empty string at the end of this program and s2 is "starting string post f call". I expected s2 to be empty as well.