From lvalue, prvalue, and xvalue to move semantics in C++

Introduction

In C++ values are divided into different categories. Knowing them helps us understand the compiler’s behind the scene actions and optimise them to our benefit.

In older than C++11, a value was either rvalue or lvalue. An lvalue was anything that could be at the left-hand side (LHS) of an assignment, =, operator. An rvalue was anything that is allowed to be on the right-hand side (RHS) of an assignment but not LHS.

int y = 5;

So in the example above y is an lvalue and 5 is an rvalue. Because a statement like 5 = y is not allowed.

In modern C++, things got a bit more complex, nowadays we have lvalue, prvalue and xvalue.

lvalue

Generally, a named variable that we can have it on the LHS of an assignment:

int y = 10;  // y is lvalue
double m = 1.5; // m is lvalue

int* p = new int{5}; // p is lvalue
*p = 4;  // *p is lvalue too

int a[5];
a[1] = 10; // array element is lvalue

int& r = y; // r is lvalue reference

In the above example, r is called lvalue reference. It is an alias for an lvalue. In other words, it binds to an lvalue. So, the RHS of an lvalue reference must be an lvalue.

(.Get 1)
int& r = 5; // Error, 5 is not an lvalue

prvalue

A prvalue or pure rvalue can only be at the RHS of an assignment.

For example, literals

int x = 2; // 2 is prvalue
string name = "Jack"; // "Jack" is prvalue
bool a = true; // true is prvalue

A temporary object is created with a literal, then it is passed to the LHS or lvalue.

RHS expressions that have an outcome due to some operations are rvalues

int x=1,y=1;
int a, b, c;
double d;

a = x+y; // the outcome of x+y is prvalue
b = -x; //  -x is prvalue
c = x+1; // x+1 is prvalue
d = (double) c; // cast is prvalue

In the above example, x+y, -x, x+1, and (double)c are calculated first which result in temporary objects. The objects are then passed to the copy assignments. The rvalue objects are destructed after assignment i.e. when we reach ; the rvalues are destructed.

The result of a function returning by value is prvalue:

int f(int i){ return i;}

int x = f(5); // the outcome of f(5) is rvalue

Objects which are created without names

class A{};
A y;
y = A{}; // the object A{} is rvalue

The list of prvalues is longer than this but a simple test to know if an RHS is prvalue is to switch LHS with RHS, and seeing whether it makes sense:

A{} = y; // doesn't make sense.
(.Get 1)

rvalue reference

An lvalue reference or simply a reference can bind to another lvalue

int x;
int& y =x;

We have the same for rvalues. An rvalue reference is defined as T&& and it only binds to rvalues

int x;

int&& y = x; // Error: cannot bind rvalue reference to lvalue

int&& y = 5; // OK: y binds to rvalue object of 5.

y = 7; // OK: y still refering to the same memory which was 5 
cout<< y; // 7

cout<< is_same<int&&, decltype(y)>::value; // true

In the previous section, I mentioned rvalues are destructed when we hit ;, but we have an exception here. The lifetime of rvalue 5 is extended to the lifetime of y due to the binding.

Note that we wrote y=7, therefore, a named rvalue reference is lvalue. To prove that see this one:

int&& y = 5;
int&& z = y; // Error: y is lvalue

We can overload a function based on a parameter being lvalue or rvalue:

void f(int& i){
    cout<< "lvalue reference called";
}

void f(int&& i){
    cout<< "rvalue reference called";
}

int x = 10;
int&& y =7;

f(x); //lvalue reference called.
f(5); //rvalue reference called.
f(y); // lvalue reference called

So x and y are lvalues, the first function is called, 5 is an rvalue so the second function is called.

(.Get 1)

move constructor: steal rvalue resource

We can improve code efficiency by making use of resources of an rvalue. I explain it with an example. We have Player class as

struct Player{
    string name;
};

We have a team class with three constructors

struct Team{
    Team(){
        goalKeeper = new Player{.name="Marc"};
    };
    Team(const Team& t){
        goalKeeper = new Player{*t.goalKeeper};
    };
    Team(Team&& t){ // move constructor
        goalKeeper = t.goalKeeper;
        t.goalKeeper = nullptr;
    };
    ~Team(){delete goalKeeper;}

   Player* goalKeeper;    
};

The first constructor is the default one. The second one constructs the object with an lvalue reference which reads the argument, t. But the third one steals the goalKeeper object of t. We are allowed to do that because the object is an rvalue, when the constructor finishes its job, t will be destructed. The third constructor is called move constructor. Similar behaviour can be defined for the assignment which is called move assignment.

Let’s implement the code:

int main(){
   Team Barca{ Team{} }; 

    return 0;
}

So, in the example above, a temporary rvalue object is created by Team{} which calls the default constructor. The object is passed to the constructor of Barca. Because it is an rvalue, the move constructor is called. The resources of the temporary object are moved to Barca. After Barca created, the temporary object is destructed. Note standard containers have built-in move constructors.

std::move

There are situations that a programmer knows that an lvalue object will be destructed soon and wants to take its resources using a move constructor/assignment. std::move casts an lvalue to an unnamed rvalue reference type. Note that std::move doesn’t move anything it is just a static cast without computational cost.

Let’s use assignment operator with rvalue reference (move assignment):

#include <iostream> 
using namespace std;

struct Player{
    string name;
};

struct Team{
    Team(){cout<<"default"<<'\n';};
    Team(const Team& t){
        cout<<"ref ctor"<<'\n';
        goalKeeper = new Player{*t.goalKeeper};
    };
    Team& operator=(const Team& t){
        cout<<"ref assign"<<'\n';
        goalKeeper = new Player{*t.goalKeeper};
        return *this;
    };
    Team& operator=(Team&& t){
        cout<<"rval assign"<<'\n';
        goalKeeper = t.goalKeeper;
        t.goalKeeper = nullptr;
        return *this;
    }
    ~Team(){
        cout<<"delete"<<'\n';
        delete goalKeeper;}

   Player* goalKeeper;    
};

Now let’s define an lvalue object and cast it to be an rvalue:

int main(){
    
   Team Barca{}; // default
   Team Real{}; // default

   {
     Team temp{}; // default
     temp.goalKeeper = new Player{.name="Marc"};

     Real = temp; // ref assign
     Barca = move(temp); // rval assign
     
   }
   // delete : temp
    return 0;
} // delete delete: Barca and Real

In the above example, temp is going out of the scope to be destructed. Before that happens, we cast temp to rvalue reference with std::move then pass it to move assignment of Barca.

xvalue

Graduating std::move, now we can define xvalue. An expiring value or xvalue is a value that is about to die so we can steal its resources. The result of a function like std::move() which returns an unnamed rvalue reference, T&&, is an xvalue:

void f(int& i){
    cout<< "lvalue reference called";
}

void f(int&& i){
    cout<< "rvalue reference called";
}

int x = 10;
f(x); // lvalue reference called
f(std::move(x)); //rvalue reference called.

A cast to an rvalue reference is an xvalue:

// using previous example functions
f(static_cast<int&&>(x)); //rvalue reference called.

An expression to access a member of an rvalue object is an xvalue:

struct A{ int i=5;};

int j = A{}.i; // A{}.i is xvalue

Where to use move semantics?

The best place to take advantage of move semantics is move constructors and assignments for classes that have movable data. In this way, we avoid the deep copy of rvalues.

However, I wouldn’t employ them in every class because the speed gain would be in assignment and constructor calls. The improvement in those actions is hardly visible if we are not moving massive objects many times. On my laptop, the deep-copy of a vector of 1 million doubles takes only 1 millisecond. Moreover, adding move constructors/assignments and std::move, st::forward and related commands makes the code harder to read and maintain. Furthermore, there are cases that a compiler itself reduces the number of objects created (Copy elision). So, if the performance gain is negligible move semantics are better to be avoided.

There are, of course, other scenarios. For example, if we write a generic library that is supposed to be used in other projects. It will be more likeable to others if the API of the library supports the move semantics.

References

cppreference.

Tags ➡ C++

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Comments

3 comments
Mike P. 19-Jan-2023
This is a great explanation. While trying some of your examples, I came across a curiosity. When you are going over named rvalue references, you provide the example:
int&& y = 5;
int&& z = y; // Error: y is lvalue

I tried this and got the expected compiler error 'You cannot bind an lvalue to an rvalue reference'. However the following code compiles fine:
int&& y = 5;
int&& z = 8;
z = y;
How could this be? z is still a named rvalue reference, hence can only bind to rvalues right? Yet y is an lvalue, further, if I had assigned y one line above, I get the error. What is happening here?
Sorush 21-Jan-2023
@Mike P., The step below
int&& y = 5; // initialization & binding
is the initialization (construction) of the reference which binds to an rvalue, in this case, 5, which is a prvalue. So the same is OK for z too:
int&& z = 8; // initialization & binding
Now both are bound and referring to unnamed memory spaces. The next steps are assignment:
z=y; // assignment
z=10; // just assignment
So in the first line above, the memory location that z is a reference of will be filled with the value of y and in the second line with 10.
This is exactly the same as lvalue references:
int i;
int j;
int& x=i; // binding to memory of i
int& y=j; // binding to memory of j
x=y; // assignment to memory of x (or i)
megorit 10-Apr-2023
String literal is an lvalue and is passed by lvalue reference.