CODEWITHSUNDEEP
c++c++11move-semantics
Stachu Jemioła
Stachu Jemioła

Reputation: 63

When move semantics works with std::move?

How does move semantics work in this example:

struct test {
    int ii[10];
    int i;
};

test f() {
    test a;
    std::cout << "[1] " << &a << std::endl;
    return a;
}

int main()
{
  test x (f());
  std::cout << "[2] " << &x << std::endl;
  test x1 (std::move(x));
  std::cout << "[3] " << &x1;
}

Output:

[1] 0x7fff50ac70c0
[2] 0x7fff50ac70c0
[3] 0x7fff50ac70f0

Why x was constructed using return value from f(), but x1 got different address than x?

Edit

struct test {
  std::string s;
}

[...]

std::cout << (*void)&x.s[0];

I think I've understood eventually. Now addresses are the same.

Upvotes: 0

Views: 94

Answers (1)

TartanLlama
TartanLlama

Reputation: 65600

This doesn't really have anything to do with move semantics. a and x have the same memory address because the copy is elided, so the return of f is allocated directly at the call site. x1 does not have the same address as x because it is a separate object, and separate objects cannot have the same address. Moving doesn't change the address, it allows the guts of your object to be ripped out and sellotaped into the moved-to object.

Upvotes: 5

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