identify some subtle effect on copy or move of std::move(lvalue reference) and const

do-not-move-lvalue-reference.cc

#include <iostream>
#include <string>

using namespace std;

struct X {
public:
    X(string s): _s(s) { cout << "x ctor\n"; }
    X(X&& x): _s(std::move(x._s))  { cout << "x move ctor\n"; } //define this will implicitly delete the copy ctor
    //X(const X& x): _s(x._s) { cout << "x copy ctor\n"; }
    void print() {
        cout << "x::print:" << _s << endl;
    }

    string _s;
};

void f1(X& x) {
    //auto& x1 = std::move(x);
    const auto x2 = std::move(x);//must use const ! This will move original data to x2

}

void f2(const X& x) {
    //auto& x1 = std::move(x);//this will not call copy ctor or move ctor
    auto x2 = std::move(x); // this will call copy ctor, so it won't affect original(Seems that compiler turn `move` into `copy`)

}

void t1() {
    X x("a");
    f1(x);

    x.print();

}

///////////////////////////////
class T {
public:
    T() { cout << "t ctor\n"; }
    T(string s) { cout << "t ctor2\n"; }
    T(T&& t) { cout << "t move ctor\n"; }
    T(const T&& t) { cout << "t copy ctor\n"; }

    string _s;
};

void g(const T& t) {
    cout << "g begin\n";
    auto t2 = std::move(t);

    cout << "------\n";
    T t1;
    T t3 = std::move(t1); // t1 is not a reference, so `move` Operation will definitely move its data to t3
}



void t2() {
    T t1;
    g(t1);
    
}

//////////////////////////////////////
void h1(const string& i) { //whether i is const& or & really matters
    string j = std::move(i);//false move, actually copy
}

void h2(string& i) {
    string j = std::move(i);//real move
}

void t_string() {
    string s = "ab";
    h1(s);
    cout << "after h1, s=" << s << endl;

    s = "abcd";
    h2(s);
    cout << "after h2, s=" << s << endl;
}

int main(int argc, char *argv[])
{
    t1();
    //t_string();
    //t2();
    return 0;
}