Why am I allowed to copy unique_ptr?

Question

Possible Duplicate:
Returning unique_ptr from functions

20.7.1.2 [unique.ptr.single] defines copy constructor like this :

// disable copy from lvalue
unique_ptr(const unique_ptr&) = delete;
unique_ptr& operator=(const unique_ptr&) = delete;

So, why the following code compiles fine?

#include <memory>
#include <iostream>

std::unique_ptr< int > bar()
{
  std::unique_ptr< int > p( new int(4));
  return p;
}

int main()
{
  auto p = bar();

  std::cout<<*p<<std::endl;
}

I compiled it like this :

g++ -O3  -Wall -Wextra -pedantic -std=c++0x kel.cpp

The compiler : g++ version 4.6.1 20110908 (Red Hat 4.6.1-9)

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.