Question

Do I really have to encapsulate the std::move call in a lambda?

std::list<std::wstring>     srcData = GetData(); // implementation not important
std::vector<std::wstring>   dstData;
dstData.reserve(srcData.size());
std::transform(std::begin(srcData), std::end(srcData), std::back_inserter(dstData), 
    [](std::wstring& guid) -> std::wstring { return std::move(guid); });
srcData.clear();

I am still new to lambdas and rvalue references, so initially I tried:

std::transform(std::begin(srcData), std::end(srcData), 
    std::back_inserter(dstData), &std::move<std::wstring>);

which doesn’t work.

Do I have to put the move inside a lambda, or am I missing something obvious?

Answer 1

An alternative is to use move iterators:

std::vector<std::wstring> dstData(std::make_move_iterator(srcData.begin()),
                                  std::make_move_iterator(srcData.end()));

Or use the move algorithm:

std::move(srcData.begin(), srcData.end(), std::back_inserter(dstData));

---

Since it was asked, here’s how you could force the original proposal to work:

int main()
{
    std::transform(std::begin(srcData),
      std::end(srcData),
      std::back_inserter(dstData),
      static_cast<std::wstring&&(*)(std::wstring&)>(&std::move<std::wstring&>));
}