vector的emplace_back

Question

Can you please explain how "perfect forwarding" works?

I read that vector's emplace_back doesn't need to copy nor move objects, because its argument is implemented as variadic template.

std::vector<T>::emplace_back(_Args&&... __args)

Can you describe it in more detail? Why won't it copy nor move?

Answer 1

emplace_back directly constructs the element at the correct position in the vector. Think of it as if

vector<T> v;
v.emplace_back(a,b,c);

is transformed into (idx being the new index)

new (v.data()+idx) T(a,b,c);

(The reality is a bit more complex involving forwarding the arguments as std::forward<_Args>()... but that might be more confusing to get the key of emplace operations)

Answer 2

There are actually two things happening in emplace_back:

1. You don't pass an object of type T but arguments to a constructor of T. This way, object construction is delayed: the vector is extended to accomodate for the memory needed by the new object, and the constructor is called to initialized the object in the vector. Variadic template don't have anything to do with copies, they only allow forwarding a variable number of arguments to the constructor.
2. Arguments to the constructors themselves are not copied because they are passed as rvalues references and std::move is used to forward them to the constructor. Basically, move semantics avoid deep copies of objects.