copy constructor is called when returning dereference of a pointer from a function

Question

I've found this weird behavior in Cpp — a copy constructor is called when a function returns dereference of a pointer. The follwing code demonstrates it. I've checked this using Visual Studio community 2022 and G++.exe.

#include <iostream>
using namespace std;

class Cls1 {
public:
    Cls1() = default;
    Cls1(Cls1& other) {
        cout << "Cls1(Cls1 & other)" << endl;
    };
    Cls1(Cls1&& other) noexcept {
        cout << "Cls1(Cls1&& other)" << endl;
    };
};

Cls1 Fn1(Cls1 v) {
    return v;
}

Cls1 Fn2(Cls1 v) {
    return *(& v); // I am mentioning this line.
}

int main(int argc, char** argv) {

    Cls1 inst1;
    Cls1 inst2 = Fn1(inst1);
    Cls1 inst3 = Fn2(inst1);

    return 0;
}

The output is the following.

Cls1(Cls1 & other)
Cls1(Cls1&& other)
Cls1(Cls1 & other)
Cls1(Cls1 & other) // I expected "Cls1(Cls1&& other)" here.

Here, Fn1 calls the move constructor and Fn2 calls the copy constructor. I was surprised because I firmly expected F2 to call the move constructor as well.

`*(& v)` is an lvalue expression https://en.cppreference.com/w/cpp/language/value_category _"...*p, the built-in indirection expression..."_ — Richard Critten, Oct 17 '22 at 07:53
A named variable is always an `lvalue` - from the above link - "_...the name of a variable... Even if the variable's type is rvalue reference, the expression consisting of its name is an lvalue expression;"_ — Richard Critten, Oct 17 '22 at 12:54
Local variables are [special](https://stackoverflow.com/q/11817873/8586227). — Davis Herring, Oct 17 '22 at 14:00

copy constructor is called when returning dereference of a pointer from a function

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