Why is this an issue?
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);
}
}
}
Resources
