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

一世执手 提交于 2019-12-06 16:05:26
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.

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;}
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