Question

I have code as following

std::vector<std::unique_ptr<int>> v;
std::unique_ptr<int> a(new int(0));
std::unique_ptr<int>& b = a;
v.insert(v.begin(), std::move(b)); //ok

However if I add const in the third statement

const std::unique_ptr<int>& b = a;
v.insert(v.begin(), std::move(b)); //Compiler error, cannot access ptr private member

Why does the compiler show it cannot access private member of the unique pointer other than cannot convert const to non-const? Thanks.

Answer 1

You cannot move a const object. Moving means altering the state of the object you moved from (of whatever type that object is), and declaring it as const makes a promise that you won’t alter that state. Thus, moving a const object is a contradiction in terms.

Although the expectation of being able to move a const object may indeed have some kind of justification (for instance, you may want the object to be const as long as you use it, and then get rid of it when you’re done and transfer its guts rather than copying it – which is btw impossible with a unique_ptr), actually allowing this would have a number of very impractical semantic consequences (for instance, it would allow swapping two const objects).

Thus, even though the language does technically allow moving from a const object (move constructors accepting a const RRef are legal), no type of the Standard Library actually gives you that possibility, including unique_ptr, and you should not write any UDT that does.

Conceptually, it is really important to think of moving as a state-altering operation, which is not compatible with the const modifier.