将右值引用绑定到（自动生成的）左值

Question

My search found many posts on rvalue binding to lvalue but not anything similar. Sorry if it is a duplicate.

struct StrHolder {
    StrHolder(std::string&& s) : name(s) {}
    void Print() const { std::cout << "My name is " << name << std::endl; }
    std::string name;
};

int main()
{
    StrHolder s{"Tom"};  // (1) - OK, as expected
    s.Print();

    std::string n1 {"Angi"};
    StrHolder p{std::move(n1)};  // (2) - OK, also as expected
    p.Print();

    //StrHolder q{n1}; // (3) - NOT OK. Also expected. Cannot bind rvalue reference to lvalue
    //q.Print();

    auto name1 {"Bon"}; // name1 is an lvalue
    StrHolder z{name1}; // (4) - Why is this OK ?
    z.Print();

    return 0;
}

The variable 'name1' declared as auto, above is an lvalue. Therefore, initialization of 'z' should fail but it does not.

Am I missing anything here ?

Answer 1

name1 is an lvalue... but it's not a std::string , it's a char const* . Constructing a StrHolder from it involves making a temporary std::string using its implicit constructor from char const* , then invoking StrHolder::StrHolder() with an rvalue reference to that temporary. name1 gets left alone, and is never moved from.