Assigning reference argument to local variable

Question

Does the following code invoke any type of implementation-defined or undefined behavior? I'm unsure how the interaction with the reference is working and my Google/SO searches are coming up empty:

struct S {
    int i;
};

void Fn(S& s_arg) {
    S s_fn{s_arg.i+1};
    s_arg = s_fn;
}

int main(int argc, char** argv) {
    S s_main{15};
    Fn(s_main);
    return 0;
}

I'm unsure which of the two should occur when the assignment occurs in Fn:

1. The default copy assignment operator of S is invoked on s_main (being the target of the s_arg reference), copying the data from the local (to Fn) s_fn into main's local s_main (via the s_arg reference, making everything proper and well-defined.
2. The reference itself is assigned and now refers to Fn's local s_fn. Fn is now returning a reference to local data and the program is now just waiting for another function to be called from main, overwriting Fn's local s_fn and causing general mayhem.

Answer 1

There is no UB here, the reference points to s_main and you are assigning to the value of s_fn to s_arg (which is pointing to s_main) and all is well. Remember that references (unlike pointers) once initialized cannot point to another region in memory which means the second case you mentioned cannot happen.