Move Semantics in C++

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This is a note for Lecture 13, CS106L, Spring 2023.

DEFINITION

L-VALUE

l-value can appear on the left or right of an =.

For example, here x is an l-value:

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int x = 3;
int y = x;
  • l-values have names
  • l-values are not temporary
  • l-values live until the end of the scope

R-VALUE

r-value can ONLY appear on the right of an =

For example, here 3 is an r-value:

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int x = 3;
int y = x;
  • r-values don't have names
  • r-values are temporary
  • r-values live until the end of the line

EXAMPLES

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int x = 3;                       // 3 is an r-value
int *ptr = 0x02248837;           // 0x02248837 is an r-value
vector<int> v1{1, 2, 3};         // {1, 2, 3} is an r-value, v1 is an l-value
auto v4 = v1 + v2;               // v1 + v2 is an r-value
size_t size = v.size();          // v.size() is an r-value
v1[1] = 4 * i;                   // 4 * i is an r-value, v1[1] is an l-value
ptr = &x;                        // &x is an r-value
v1[2] = *ptr;                    // *ptr is an l-value
MyClass obj;                     // obj is an l-value
x = obj.public_member_variable;  // obj.public_member_variable is l-value

MOVE SEMANTICS

In our generic vector class, we have a vector copy assignment operator like this:

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template <typename T>
vector<T>& vector<T>::operator=(const vector<T>& other) {
  if (&other == this) return *this;
  _size = other._size;
  _capacity = other._capacity;
  delete[] _elems;
  _elems = new T[other._capacity];
  std::copy(other._elems, other._elems + other._size, _elems);
  return *this;
}

Aside: std::copy is a generic copy function used to copy a range of elements from one container to another.

And in the code fragment below:

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int main() {
  vector<int> vec;
  vec = make_me_a_vec(123);
}
  • vec is created using the default constructor
  • make_me_a_vec creates a vector using the default constructor
  • vec is reassigned to a copy of that return value using copy assignment
  • copy assignment creates a new array and copies the contents of the old one
  • The original return value's lifetime ends and it calls its destructor
  • vec's lifetime ends and it calls its destructor

Here is a problem: make_me_a_vec(123) is an r-value, and in vector<T>::operator=(const vector<T>& other), other should be an l-value (referenced using &). Can r-values be bound to const &?

The answer is Yes.

Another problem is that, we creates a vector, copies its content to another and deleted it. Can we do better?

We can use move assignment like this:

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template <typename T>
vector<T>& vector<T>::operator=(const vector<T>& other) {
  if (&other == this) return *this;
  _size = other._size;
  _capacity = other._capacity;
  _elems = other._elems; // we don't copy in this version
  return *this;
}

But what about this?

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int main() {
  vector<string> vec1 = {"hello", "world"};
  vector<string> vec2;
  vec2 = vec1;
  vec1.push_back("Sure hope vec2 doesn't see this!");
} // BAD!

A problem occurs here!We need both a copy assignment AND a move assignment.

How do we know when to use move assignment and when to use copy assignment?

When the item on the right of the = is an r-value we should use move assignment.

Why? r-values are always about to die, so we can steal their resources.

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int main() {
  vector<int> vec;
  vec = make_me_a_vec(123); // using move assignment
}
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int main() {
  vector<string> vec1 = {"hello", "world"};
  vector<string> vec2;
  vec2 = vec1; // using copy assignment
  vec1.push_back("Sure hope vec2 doesn't see this!");
} // and vec2 never saw a thing

And now the question is: how to make two different assignment operator?

Answer: Overload vector::operator=!

Introducing... the r-value reference using &&

R-VALUE REFERENCE

By using r-value reference, we can do this:

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int main() {
  int x = 1;
  change(x); // this will call version 2
  change(7); // this will call version 1
}
void change(int&& num) {...} // version 1 takes r-values
void change(int& num) {...}  // version 2 takes l-values
// num is a reference to int

So, we should keep our copy assignment:

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vector<T>& operator=(const vector<T>& other) {
  if (&other == this) return *this;
  _size = other._size;
  _capacity = other._capacity;

  // must **copy** entire array
  delete[] _elems;
  _elems = new T[other._capacity];
  std::copy(other._elems, other._elems + other.size, _elems);
  return *this;
}

And overload vector::operator= (move assignment) like this:

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vector<T>& operator=(vector<T>&& other) {
  if (&other == this) return *this;
  _size = other._size;
  _capacity = other._capacity;

  // we can **steal** the array
  delete[] _elems;
  _elems = other._elems;
  return *this;
}

But actually, we still copy _size and _capacity, etc.

Introducing...std::move!

std::move

  • std::move(x) doesn't do anything except cast x as an r-value
  • It is a way to force C++ to choose the && version of a function
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int main() {
  int x = 1;
  change(x);             // this will call version 2
  change(std::move(x));  // this will call version 1
}
void change(int&& num) {...} // version 1 takes r-values
void change(int& num) {...}  // version 2 takes l-values

We can modify our move assignment like this:

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vector<T>& operator=(vector<T>&& other) {
  if (&other == this) return *this;
  _size = std::move(other._size);          // force move assignment
  _capacity = std::move(other._capacity);  // force move assignment

  // we can **steal** the array
  delete[] _elems;
  _elems = std::move(other._elems);        // force move assignment
  return *this;
}

This works!

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int main() {
  vector<int> vec;
  vec = make_me_a_vec(123); // this will use move assignment
  vector<string> vec1 = {"hello", "world"};
  vector<string> vec2;
  vec2 = vec1; // this will use copy assignment
  vec1.push_back("Sure hope vec2 doesn't see this!");
}

However, what if we wanted to declare and initialize a vec on the same line?

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int main() {
  vector<int> vec;
  vec = make_me_a_vec(123); // this will use move assignment
  vector<string> vec1 = {"hello", "world"};
  vector<string> vec2 = vec1; // this will use copy constructor
  vec1.push_back("Sure hope vec2 doesn't see this!");
}

Similarly, vector<string> vec1 = {"hello", "world"}; will use move constructor.

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vector<T>(vector<T>&& other) {
  if (&other == this) return *this;
  _size = std::move(other._size);
  _capacity = std::move(other._capacity);

  // we can steal the array
  delete[] _elems;
  _elems = std::move(other._elems);
  return *this;
}

Where else should we use std::move?

Rule of Thumb:
Wherever we take in a const & parameter in a class member function and assign it to something else in our function
(TO BE CONTINUED)

For example:

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// copy push_back
void push_back(const T& element) {
  elems[_size++] = element;
  // this is copy assignment
}

// move push_back
void push_back(T&& element) {
  elems[_size++] = std::move(element);
  // this forces T's move assignment
}

Be careful with std::move

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int main() {
  vector<string> vec1 = {"hello", "world"};
  vector<string> vec2 = std::move(vec1);
  vec1.push_back("Sure hope vec2 doesn't see this!"); // WRONG
}

Rule of Thumb:
Wherever we take in a const & parameter in a class member function and assign it to something else in our function
Don't use std::move outside of class definitions, never use it in application code!

TLDR

  • If your class has copy constructor and copy assignment defined, you should also define a move constructor and move assignment

  • Define these by overloading your copy constructor and assignment to be defined for Type&& other as well as Type& other

  • Use std::move to force the use of other types' move assignments and constructors

  • All std::move(x) does is cast x as an r-value

  • By wary of std::move(x) in main function code!

PHILOSOPHY about SMFs

The 6 Special Member Functions

  1. Default constructor: Initializes an object to a default state
  2. Copy constructor: Creates a new object by copying an existing object
  3. Move constructor: Creates a new object by moving the resources of an existing object
  4. Copy Assignment Operator: Assigns the contents of one object to another object
  5. Move Assignment Operator: Moves the resources of one object to another object
  6. Destructor: Frees any dynamically allocated resources owned by an object when it is destroyed

Some Philosophy about SMFs

There are three guiding rules:

Rule of Zero

  • If you can avoid defining default operations, do
  • Why? It's the simplest and gives the cleanest semantic
  • Example: Since std::map and std::string have all the special functions, no further work is needed.
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Class Named_map {
 public:
  // ...no default operations declared
 private:
  std::string name;
  std::map<int, int> rep;
};

Named_map nm;      // default construct
Named_map nm2{nm}; // copy construct

Rule of Three

  • If you need to implement a custom destructor, you almost certainly need to define a copy constructor and copy assignment operator
  • Why? You are probably managing your own memory somehow, so the shallow copies provided by the default operations won't work correctly

Rule of Five

  • If you define custom copy constructor/assignment operator, you should define move constructor/assignment operator as well
  • Why? This is about efficiency rather than correctness. It's inefficient to make extra copies (although it's "correct")