Why am I allowed to copy unique_ptr? [duplicate]

Answer 1

In the return statement, if you return a local variable, the expression is treated as an rvalue, and thus automatically moved. It is thus similar to:

  return std::move(p);

It invokes the unique_ptr(unique_ptr&&) constructor.

In the main function, bar() produces a temporary, which is an rvalue, and is also properly moved into the p in main.

Answer 2

It is not copied, it is moved.

The return statement is equivalent to this:

return std::move(p);

Pedantically speaking, that is semantically equivalent. In reality, the compiler may optimize the code, eliding the call to the move-constructor. But that is possible only if you write it as:

return p; //It gives the compiler an opportunity to optimize this. 

That is recommended. However, the compiler has no opportunity to optimize if you write this:

return std::move(p); //No (or less) opportunity to optimize this. 

That is not recommended. :-)

Answer 3

I think that copying from an lvalue is disabled, but "bar()" is an rvalue so it's OK. You definitely need to be able to copy from rvalues.