Why am I allowed to copy unique_ptr? [duplicate]

问题

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)

回答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.

回答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. :-)

回答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.

来源：https://stackoverflow.com/questions/9827183/why-am-i-allowed-to-copy-unique-ptr

标签

c++

g++

c++11

unique-ptr