If you use ``++std::unique_ptr<T> const &++`` for a function parameter type, it means that the function will not be able to alter the ownership of the pointed-to object by the ``++unique_ptr++``:

* It cannot acquire ownership of the pointed-to object (this would require a parameter of type ``++std::unique_ptr<T>++``)
* It cannot transfer the object ownership to someone else (this would require a ``++std::unique_ptr<T> &++``).

That means the function can only observe the pointed-to object, and in this case passing a ``++T*++`` (if the ``++unique_ptr++`` can be null) or a ``++T&++`` (if it cannot) provides the same features, while also allowing the function to work with objects that are not handled by a ``++unique_ptr++`` (E.G. objects on the stack, in a ``++vector++``, or in another kind of smart pointer), thus making the function more general-purpose.


== Noncompliant Code Example

----
using namespace std;
void draw(unique_ptr<Shape> const &shape); // Noncompliant

void drawAll(vector<unique_ptr<Shape>> v)
{
  for (auto &shape : v) {
      if (shape) {
        draw(shape);
      }
  }
}
----


== Compliant Solution

----
using namespace std;
void draw(Shape const &shape); // Compliant

void drawAll(vector<unique_ptr<Shape>> v)
{
  for (auto &shape : v) {
      if (shape) {
        draw(*shape);
      }
  }
}
----


== See

* https://github.com/isocpp/CppCoreGuidelines/blob/036324/CppCoreGuidelines.md#r32-take-a-unique_ptrwidget-parameter-to-express-that-a-function-assumes-ownership-of-a-widget[{cpp} Core Guidelines R.32] - Take a unique_ptr<widget> parameter to express that a function assumes ownership of a widget

