copy vs std::move for ints

问题:

• What's difference between default copy and std::move in that example?
• After move the object is there any dependence between new and old ones?

int main () {

    int a = 100;
    std::cout<<&a<<std::endl;

    auto a_copy = a;                 // deduced as int
    std::cout<<&a_copy<<std::endl;

    auto a_move = std::move(a);      // deduced as int
    std::cout<<&a_move<<std::endl;

};

output:

0x7fffffffe094
0x7fffffffe098
0x7fffffffe09c

回答1:

In this example, there is no difference. We will end up with 3 ints with value 100. There could definitely be a difference with different types though. For instance, let's consider something like vector<int>:

std::vector<int> a = {1, 2, 3, 4, 5}; // a has size 5
auto a_copy = a;                      // copy a. now we have two vectors of size 5
auto a_move = std::move(a);           // *move* a into a_move

The last variable, a_move, takes ownership of a's internal pointers. So what we end up with is a_move is a vector of size 5, but a is now empty. The move is much more efficient than a copy (imagine if it was a vector of 1000 strings instead - a_copy would involve allocating a 1000-string buffer and copying 1000 strings, but a_move just assigns a couple pointers).

For some other types, one might be invalid:

std::unique_ptr<int> a{new int 42};
auto a_copy = a;            // error
auto a_move = std::move(a); // OK, now a_move owns 42, but a points to nothing

For many types, there's no difference though:

std::array<int, 100> a;
auto a_copy = a;            // copy 100 ints
auto a_move = std::move(a); // also copy 100 ints, no special move ctor

More generally:

T a;
auto a_copy = a;            // calls T(const T& ), the copy constructor
auto a_move = std::move(a); // calls T(T&& ), the move constructor

回答2:

Using std::move just changes an lvalue to an xvalue, so it is eligible for use with move constructors and move assignment operators. These do not exist for built-in types, so using move makes no difference in this example.

回答3:

What's difference between default copy and std::move in that example?

There is no difference. Copying something satisfies the requirements of a move, and in the case of built-in types, move is implemented as a copy.

After move the object is there any dependence between new and old

No, there are no dependencies. Both variables are independent.

回答4:

To expand on the other poster's answer, the move-is-a-copy paradigm applies to all data structures composed of POD types (or composed of other types composed of POD types) as well, as in this example:

struct Foo
{
    int values[100];
    bool flagA;
    bool flagB;
};

struct Bar
{
    Foo foo1;
    Foo foo2;
};

int main()
{
    Foo f;
    Foo fCopy = std::move(f);
    Bar b;
    Bar bCopy = std::move(b);
    return 0;
}

In the case of both Foo and Bar there is no meaningful way to move the data from one to another because both are ultimately aggregates of POD types - none of their data is indirectly owned (points to or references other memory). So in these cases, the move is implemented as a copy and originals (f, b) remain unaltered after the assignments on the std::move() lines.

Move semantics can only be meaningfully implemented with dynamically allocated memory or unique resources.