std::unique_ptr<T,Deleter>::operator= - cppreference.com (original) (raw)

1. Move assignment operator. Transfers ownership from r to *this as if by calling reset(r.release()) followed by assigning get_deleter() from std::forward<Deleter>(r.get_deleter()).

This overload participates in overload resolution only if std::is_move_assignable<Deleter>::value is true.

If Deleter is not a reference type, the behavior is undefined if

• Deleter is not MoveAssignable, or
• assigning get_deleter() from an rvalue of type Deleter would throw an exception.

Otherwise (Deleter is a reference type), the behavior is undefined if

• std::remove_reference<Deleter>::type is not CopyAssignable, or
• assigning get_deleter() from an lvalue of type Deleter would throw an exception.

1. Converting assignment operator. Transfers ownership from r to *this as if by calling reset(r.release()) followed by assigning get_deleter() from std::forward<E>(r.get_deleter()).

This overload participates in overload resolution only if all following conditions are satisfied:

• std::is_assignable<Deleter&, E&&>::value is true.
• For the primary template, all following conditions are satisfied:
  • U is not an array type.
  • unique_ptr<U, E>::pointer is implicitly convertible to pointer, and.
• For the array specialization (unique_ptr<T[]>), all following conditions are satisfied:
  • U is an array type.
  • pointer is the same type as element_type*.
  • unique_ptr<U, E>::pointer is the same type as unique_ptr<U, E>::element_type*.
  • unique_ptr<U, E>::element_type(*)[] is convertible to element_type(*)[].

If E is not a reference type, the behavior is undefined if assigning get_deleter() from an rvalue of type E is ill-formed or would throw an exception.

Otherwise (E is a reference type), the behavior is undefined if assigning get_deleter() from an lvalue of type E is ill-formed or would throw an exception.

1. Effectively the same as calling reset().

2. Copy assignment operator is explicitly deleted.

[edit] Parameters

[edit] Return value

*this

[edit] Notes

As a move-only type, unique_ptr's assignment operator only accepts rvalues arguments (e.g. the result of std::make_unique or a std::move'd unique_ptr variable).

[edit] Example

#include #include   struct Foo { int id; Foo(int id) : id(id) { std::cout << "Foo " << id << '\n'; } Foo() { std::cout << "Foo " << id << '\n'; } };   int main() { std::unique_ptr p1(std::make_unique(1));   { std::cout << "Creating new Foo...\n"; std::unique_ptr p2(std::make_unique(2)); // p1 = p2; // Error ! can't copy unique_ptr p1 = std::move(p2); std::cout << "About to leave inner block...\n";   // Foo instance will continue to live, // despite p2 going out of scope }   std::cout << "About to leave program...\n"; }

Output:

Foo 1 Creating new Foo... Foo 2 ~Foo 1 About to leave inner block... About to leave program... ~Foo 2

[edit] Defect reports

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.