copy vs std::move for ints

Question

• 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?

Answer 1

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.