Why Does Returning a `std::unique_ptr` Not Require `std::move()`?

Why Return a std::unique_ptr Without std::move()?

std::unique_ptr prevents copy construction and instead employs move semantics. However, it's possible to return a unique_ptr from a function without explicitly using std::move().

Consider the following code snippet:

unique_ptr<int> foo()
{
  unique_ptr<int> p(new int(10));

  return p; // Line 1
}

int main()
{
  unique_ptr<int> p = foo();
  cout << *p << endl;
}

Line 1 returns a unique_ptr without std::move(). Surprisingly, the code compiles and operates as intended, so how is this possible without invoking the copy constructor?

Exploiting Language Specification

The key to understanding this behavior lies in the C language specification. Specifically, section 12.8 §34 and §35 describe copy elision:

When certain criteria are met, an implementation is allowed to omit the copy/move construction of a class object [...].

In this case, copy elision is permitted because:

• The return value is a temporary object.
• The temporary object has the same type as the function return type.

When the criteria for elision of a copy operation are met and the object to be copied is designated by an lvalue, overload resolution to select the constructor for the copy is first performed as if the object were designated by an rvalue.

This implies that although the return value is a named object (an lvalue), overload resolution still considers it as an rvalue. Thus, the move constructor is not invoked, and the returned unique_ptr is successfully constructed.

Conclusion

The ability to return a unique_ptr without std::move() is a subtle feature of the C language specification. It is not necessary to explicitly use std::move() when returning a unique_ptr from a function, as long as the conditions for copy elision are met. This can help improve code clarity and reduce unnecessary move operations.

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