关于 std::move 的一些事情

Question

std::string str1 = "hello";
std::string &&str2 = std::move(str1);
std::string str3(str2);
std::cout << str1 << std::endl;

I think at line 3 that str3 will steal str1, because std::string(std::string &&str) (or sth. like that) will be invoked. But the result is:

test_move()
hello
test_move()

I want to know why the content str1 is still there after line 3.

std::string str1 = "hello";
std::string &&str2 = std::move(str1);
std::string str3(str2);
std::cout << str1 << std::endl;

std::string str4(std::move(str1));
std::cout << str1 << std::endl;

I try to move str1 directly to str4, and the result is:

test_move()
hello 

test_move()

So I can't figure out the difference between the situation of str4 and str3. In my eyes they are all the same, all takes a rvalue ref as parameter, but move semantics takes place only on str4.Why?

Answer 1

std::move doesn't actually move anything. It merely creates an rvalue reference.

But since you initialize a variable with this rvalue reference, the variable itself becomes an lvalue. Passing the actual rvalue to an rvalue-constructor (as done in the second example) the constructor does the "moving".

If we create a small example program:

#include <iostream>
#include <utility>

struct Foo
{
    Foo() = default;

    Foo(Foo const&)
    {
        std::cout << "l-value copy construction\n";
    }

    Foo(Foo&&)
    {
        std::cout << "r-value move construction\n";
    }
};

int main()
{
    {
        Foo a;
        Foo&& b = std::move(a);
        Foo c(b);
    }

    {
        Foo a;
        Foo b(std::move(a));
    }
}

If you build and run it then you will see that it will invoke the lvalue copy constructor for your first case.

See it running eg here