如何在std :: unorderd_map中就地构造某个类的对象？

Question

I want to store a number of objects MyItem , derived from Qt::QGraphicsItem , in an std::unordered_map . As I understood, Qt::QGraphicsItem is none-copyable : copy constructor is private.

This is fine; I don't want to copy MyItem . But I would need to construct MyItem in-place. Assuming MyItem has this constructor signature:

MyItem::MyItem(int a, double b, std::string c);

And this is my std::unordered_map :

std::unordered_map< KeyType, MyItem > myItemMap;

What is wrong with this line, where KeyType is copyable and constructed before the emplacement:

myItemMap.emplace( correspondingKey, MyItem(3, 3.14, "hello") );

I get this error (cut):

use of deleted function 'MyItem::MyItem(const MyItem&)'
  : first(std::forward<_U1>(__x)), second(__y) { }
                                             ^
'MyItem::MyItem(const MyItem&)' is implicitly deleted because the default definition would be ill-formed:

'QGraphicsItem::QGraphicsItem(const QGraphicsItem&)' is private

How does the arguments of myItemMap.emplace(...) must look like?

Answer 1

You're constructing a MyItem value which emplace then tries to copy. The point of the emplace functions is to leave the construction to the map . It gets complicated because you need to pass a pair , where you want to defer construction of second .

myItemMap.emplace(std::piecewise_construct,
          std::forward_as_tuple(correspondingKey),
          std::forward_as_tuple(3, 3.14, "hello"));

See http://en.cppreference.com/w/cpp/container/unordered_map/emplace .

Answer 2

You code will work if you add a move constructor to MyItem :

MyItem(MyItem&& old)
    : Base(old.parentItem())
{
    for (QGraphicsItem* child : old.childItems()) {
        child->setParentItem(this);
    }
    old.setParentItem(nullptr);
}

Then the instance will be automatically moved and not copied.