[corysama]

Edit: Ignore this comment. I shouldn't talk so much about something I use so little. Somehow I skimmed through http://en.cppreference.com/w/cpp/memory/shared_ptr/shared_pt... and thought that shared_ptr( const shared_ptr& r ); was missing.

>> That's a good point. If the function will eventually lead to some object retaining a reference, then you should use a non-const shared_ptr ref. The assignment into the object will do the copy constructor and that needs a non-const ref. But, you don't need to be making temp object copies along the way. If the function will not lead to something retaining a reference, then you shouldn't be passing the reference retention object to it. Just pass a direct ref to the target object.

[tomjakubowski]

> The assignment into the object will do the copy constructor and that needs a non-const ref.

Hum? What assignment into the object? Where I'd typically see taking a `const std::shared_ptr<> &` to signal "retaining ownership" would be something like this:

    class Foo {
    public:
        void AppendChild(const std::shared_ptr<Foo> &x) {
            children_.emplace_back(x);
        }
    private:
        std::vector<std::shared_ptr<Foo>> children_;
    };

Why should Foo:;AppendChild's signature be changed to a non-const ref?

[millstone]

It should be changed to pass by value, and then use std::move to move it into the children array. The reason is that if the argument is a temporary, you are making an unnecessary copy of that temporary, when you could just transfer ownership.

This also gives callers the flexibility to std::move into your argument, transferring ownership.

A good rule of thumb is, if you are going to unconditionally take ownership of an object, accept it by value.

[corysama]

You are correct. I had a brain fart. Edited my comment.