std::vector initialization move/copy constructor of the element

问题

I have this piece of code:

#include <iostream>
#include <vector>

using namespace std;

class Foo{
public:
    Foo() noexcept {cout << "ctor" << endl;}
    Foo(const Foo&) noexcept {cout << "copy ctor" << endl;}
    Foo(Foo&&) noexcept {cout << "move ctor" << endl;}

    Foo& operator=(Foo&&) noexcept {cout << "move assn" << endl; return *this;}
    Foo& operator=(const Foo&) noexcept {cout << "copy assn" << endl; return *this;}

    ~Foo() noexcept {cout << "dtor" << endl;}
};


int main()
{   
    Foo foo;

    vector<Foo> v;
    v.push_back(std::move(foo)); 

    // comment the above 2 lines and replace by
    // vector<Foo> v{std::move(foo)}; 
}

The output is what I expect (compiled with g++ -std=c++11 --no-elide-constructors, same output without the flag)

ctor
move ctor
dtor
dtor

Now instead of using push_back initialize directly the vector v as

vector<Foo> v{std::move(foo)};

I do not understand why I get the outputs:

1) (without --no-elide-constructors)

ctor
move ctor
copy ctor
dtor
dtor
dtor

2) (with --no-elide-constructors)

ctor
move ctor
move ctor
copy ctor
dtor
dtor
dtor
dtor

In the first case, why is the copy ctor invoked? And in the second case, when the compiler does not perform elision, I have absolutely no idea why the move ctor is invoked twice. Any ideas?

回答1:

vector<Foo> v{std::move(foo)};

Here you're calling the vector constructor that takes an std::initializer_list. An initializer list only allows const access to its elements, so the vector is going to have to copy each element from the initializer_list to its own storage. That's what causes the call to the copy constructor.

From §8.5.4/5 [dcl.init.list]

An object of type std::initializer_list<E> is constructed from an initializer list as if the implementation allocated a temporary array of N elements of type const E, where N is the number of elements in the initializer list.

---

As for the extra move constructor call with -fno-elide-constructors, this was discussed in another answer a couple of days ago. It seems as though g++ takes a very literal approach to the example implementation of an initializer_list shown in the standard in same section I've quoted above.

The same example, when compiled using clang, doesn't produce the extra move constructor call.

回答2:

The containers try very hard to make sure they remain usable should an exception occur. As part of this, they'll only use std::move internally if your class' move constructor is exception safe. If it is not, (or it can't tell), it will copy just to be safe.

The correct move operations are

Foo(Foo&&) noexcept {cout << "move ctor" << endl;}
Foo& operator=(Foo&&) noexcept {cout << "move assn" << endl;}

来源：https://stackoverflow.com/questions/24793019/stdvector-initialization-move-copy-constructor-of-the-element