If you’re not keeping the parameter, then you still want to have separate T const& and T&& overloads

Raymond Chen

Raymond

Last time, I noted that if you plan on keeping the parameter anyway, then there’s no need to have separate T const& and T&& overloads. However, the converse also applies: If you’re not keeping the parameter, then you still want to have separate T const& and T&& overloads.

To recap, we started with a class like this:

class Widget
{
public:
  void SetValues(std::vector<int> const& values)
  {
    m_values = values;
  }

  void SetValues(std::vector<int>&& values)
  {
    m_values = std::move(values);
  }
private:
  std::vector<int> m_values;
};

We were able to simplify this to

class Widget
{
public:
  void SetValues(std::vector<int> values)
  {
    m_values = std::move(values);
  }

private:
  std::vector<int> m_values;
};

because we are going to keep the parameter either way. (The old way resulted in either a copy or a move. The new way produces either a copy+move or a move. The expectation is that a single move is relatively inexpensive.)

However, the simplification doesn’t apply if we are not the ones consuming the value.

Widget CreateWidgetWithValues(std::vector<int> values)
{
  Widget widget;
  widget.SetValues(std::move(values));
  return widget;
}

In this case, we are moving the values onward to the SetValues method, who is the final consumer. Writing the method this way generates an extra move constructor, because we have to move the value from our inbound parameter into the outbound parameter to SetValues. We also incur an extra destruction of our now-empty inbound parameter. If the parameter is passed through multiple layers, each layer adds an extra move constructor and destruction.

Since we are not the final consumer, we should forward the parameter.

template<typename Values>
Widget CreateWidgetWithValues(Values&& values)
{
  Widget widget;
  widget.SetValues(std::forward<Values>(values));
  return widget;
}

Unfortunately, this causes us to break existing code, since you cannot forward uniform initialization.

// doesn't work any more
CreateWidgetWithValues({ range.begin(), range.end() });

We end up returning to the overload.

Widget CreateWidgetWithValues(const std::vector<int>& values)
{
  Widget widget;
  widget.SetValues(values);
  return widget;
}

Widget CreateWidgetWithValues(std::vector<int>&& values)
{
  Widget widget;
  widget.SetValues(std::move(values));
  return widget;
}

I’m not too happy with this, though. Maybe there’s an easier way. Let me know.

Bonus chatter: The Microsoft compiler makes the function responsible for destructing its inbound parameters, in which case the code to destruct the std::vector<int> is part of the consuming function and is therefore shared. gcc and clang make it the caller’s responsibility, so the destruction of the parameter is repeated at each call site.

4 comments

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  • Avatar
    Zachary Turner

    I think the issue here isn’t that “you cannot forward uniform initialization”, it’s that you can’t even use uniform initialization in the first place. Or at least, not in a deduced context, because `{a.begin(), b.begin()};` could mean anything. is it a list, a vector, a map, a `Foo`? If you spell this out at the call site, I think the uniform initialization syntax works just fine. Does `CreateWidgetWithValues(std::vector{ range.begin(), range.end() });` work and allow you to have only a single implementation?

  • Avatar
    Matteo Italia

    Generally I just take by copy even in this case. Move constructors and destructors for moved-out objects are supposed to be dirty cheap; ultimately looking at the generated code the worse that happens is a bit of code bloat for code paths that aren’t “really” taken (e.g. a branch + a call to `delete` that won’t ever be invoked), but at runtime the difference is not really visible, and at compile-time you avoid duplication and keep the fact that you aren’t the “final” destination of the passed-in `std::vector` (or whatever) as an implementation detail.