Can I Return a `std::unique_ptr` Without `std::move` in C ?

Returning a std::unique_ptr without std::move

In C , std::unique_ptr enforces move semantics to prevent multiple ownership of objects. However, it's possible to return a unique_ptr from a function without invoking std::move. This behavior can be attributed to a language feature called copy elision.

Copy Elision

According to the C language specification (12.8 §34 and §35), an implementation is permitted to elide (omit) copy/move operations when certain criteria are met. Specifically, copy elision is allowed in a return statement for a class return type, provided that the expression is the name of a non-volatile automatic object with the same cv-unqualified type as the function return type.

Elision in unique_ptr Return Statements

In the example code you provided:

unique_ptr<int> foo()
{
  unique_ptr<int> p(new int(10));
  return p;  // Line 1
}</int></int>

p is a non-volatile automatic object that has the same type as the function return type, unique_ptr. Therefore, the compiler is permitted to elide the copy construction of p.

When the compiler encounters line 1, it first performs overload resolution to select the constructor for the copy operation, as if p were designated by an rvalue. However, since the criteria for elision are met, the copy construction is elided and the returned value becomes a moved-out unique_ptr.

Implications for best practices

It's important to note that returning by value should be the default choice in this scenario. In the presence of copy elision, a named value in the return statement is treated as an rvalue. This means that even without explicitly using std::move, the returned unique_ptr can be moved from the function's temporary object.

Returning by reference or using std::make_unique are both viable alternatives to avoid copy elision and ensure explicit ownership transfer.

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