What optimization does move semantics provide if we already have RVO? - c++

As far as I understand one of the purposes of adding move semantics is to optimize code by calling special constructor for copying "temporary" objects. For example, in this answer we see that it can be used to optimize such string a = x + y stuff. Because x+y is an rvalue expression, instead of deep copying we can copy only the pointer to the string and the size of the string. But as we know, modern compilers support return value optimization, so without using move semantics our code will not call the copy constructor at all.
To prove it I write this code:
#include <iostream>
struct stuff
{
int x;
stuff(int x_):x(x_){}
stuff(const stuff & g):x(g.x)
{
std::cout<<"copy"<<std::endl;
}
};
stuff operator+(const stuff& lhs,const stuff& rhs)
{
stuff g(lhs.x+rhs.x);
return g;
}
int main()
{
stuff a(5),b(7);
stuff c = a+b;
}
And after executing it in VC++2010 and g++ in optimize mode I'm getting empty output.
What kind of optimization is it, if without it my code still works faster? Could you explain what I'm understanding wrong?

Move semantics should not be thought as an optimization device, even if they can be used as such.
If you are going to want copies of objects (either function parameters or return values), then RVO and copy elision will do the job when they can. Move semantics can help, but are more powerful than that.
Move semantics are handy when you want to do something different whether the passed object is a temporary (it then binds to a rvalue reference) or a "standard" object with a name (a so called const lvalue). If you want for instance to steal the resources of a temporary object, then you want move semantics (example: you can steal the contents a std::unique_ptr points to).
Move semantics allow you to return non copyable objects from functions, which is not possible with the current standard. Also, non copyable objects can be put inside other objects, and those objects will automatically be movable if the contained objects are.
Non copyable objects are great, since they don't force you to implement an error-prone copy constructor. A lot of the time, copy semantics do not really make sense, but move semantics do (think about it).
This also enables you to use movable std::vector<T> classes even if T is non copyable. The std::unique_ptr class template is also a great tool when dealing with non copyable objects (eg. polymorphic objects).

After some digging I find this excellent example of optimization with rvalue references inStroustrup's FAQ .
Yes, swap function:
template<class T>
void swap(T& a, T& b) // "perfect swap" (almost)
{
T tmp = move(a); // could invalidate a
a = move(b); // could invalidate b
b = move(tmp); // could invalidate tmp
}
This will generate optimized code for any kind of types (assuming, that it have move constructor).
Edit: Also RVO can't optimize something like this(at least on my compiler):
stuff func(const stuff& st)
{
if(st.x>0)
{
stuff ret(2*st.x);
return ret;
}
else
{
stuff ret2(-2*st.x);
return ret2;
}
}
This function always calls copy constructor (checked with VC++). And if our class can be moved faster, than with move constructor we will have optimization.

Imagine your stuff was a class with heap allocated memory like a string, and that it had the notion of capacity. Give it a operator+= that will grow the capacity geometrically. In C++03 this might look like:
#include <iostream>
#include <algorithm>
struct stuff
{
int size;
int cap;
stuff(int size_):size(size_)
{
cap = size;
if (cap > 0)
std::cout <<"allocating " << cap <<std::endl;
}
stuff(const stuff & g):size(g.size), cap(g.cap)
{
if (cap > 0)
std::cout <<"allocating " << cap <<std::endl;
}
~stuff()
{
if (cap > 0)
std::cout << "deallocating " << cap << '\n';
}
stuff& operator+=(const stuff& y)
{
if (cap < size+y.size)
{
if (cap > 0)
std::cout << "deallocating " << cap << '\n';
cap = std::max(2*cap, size+y.size);
std::cout <<"allocating " << cap <<std::endl;
}
size += y.size;
return *this;
}
};
stuff operator+(const stuff& lhs,const stuff& rhs)
{
stuff g(lhs.size + rhs.size);
return g;
}
Also imagine you want to add more than just two stuff's at a time:
int main()
{
stuff a(11),b(9),c(7),d(5);
std::cout << "start addition\n\n";
stuff e = a+b+c+d;
std::cout << "\nend addition\n";
}
For me this prints out:
allocating 11
allocating 9
allocating 7
allocating 5
start addition
allocating 20
allocating 27
allocating 32
deallocating 27
deallocating 20
end addition
deallocating 32
deallocating 5
deallocating 7
deallocating 9
deallocating 11
I count 3 allocations and 2 deallocations to compute:
stuff e = a+b+c+d;
Now add move semantics:
stuff(stuff&& g):size(g.size), cap(g.cap)
{
g.cap = 0;
g.size = 0;
}
...
stuff operator+(stuff&& lhs,const stuff& rhs)
{
return std::move(lhs += rhs);
}
Running again I get:
allocating 11
allocating 9
allocating 7
allocating 5
start addition
allocating 20
deallocating 20
allocating 40
end addition
deallocating 40
deallocating 5
deallocating 7
deallocating 9
deallocating 11
I'm now down to 2 allocations and 1 deallocations. That translates to faster code.

There are many places some of which are mentioned in other answers.
One big one is that when resizing a std::vector it will move move-aware objects from the old memory location to the new one rather than copy and destroy the original.
Additionally rvalue references allow the concept of movable types, this is a semantic difference and not just an optimization. unique_ptr wasn't possible in C++03 which is why we had the abomination of auto_ptr.

Just because this particular case is already covered by an existing optimization does not mean that other cases don't exist where r-value references are helpful.
Move construction allows optimization even when the temporary is returned from a function which cannot be inlined (perhaps it's a virtual call, or through a function pointer).

Your posted example only takes const lvalue references and so explicitly cannot have move semantics applied to it, as there is not a single rvalue reference in there. How can move semantics make your code faster when you implemented a type without rvalue references?
In addition, your code is already covered by RVO and NRVO. Move semantics apply to far, far more situations than those two do.

This line calls the first constructor.
stuff a(5),b(7);
Plus operator is called using explicit common lvalue references.
stuff c = a + b;
Inside operator overload method, you have no copy constructor called.
Again, the first constructor is called only.
stuff g(lhs.x+rhs.x);
assigment is made with RVO, so no copy is need. NO copy from returned object to 'c' is need.
stuff c = a+b;
Due no std::cout reference, compiler take care about your c value is never used. Then, whole program is stripped out, resulting in a empty program.

Another good example I can think of. Imagine that you're implementing a matrix library and write an algorithm which takes two matrices and outputs another one:
Matrix MyAlgorithm(Matrix U, Matrix V)
{
Transform(U); //doesn't matter what this actually does, but it modifies U
Transform(V);
return U*V;
}
Note that you can't pass U and V by const reference, because the algorithm tweaks them. You could theoretically pass them by reference, but this would look gross and leave U and V in some intermediate state (since you call Transform(U)), which may not make any sense to the caller, or just not make any mathematical sense at all, since it's just one of the internal algorithm transformations. The code looks much cleaner if you just pass them by value and use move semantics if you are not going to use U and V after calling this function:
Matrix u, v;
...
Matrix w = MyAlgorithm(u, v); //slow, but will preserve u and v
Matrix w = MyAlgorithm(move(u), move(v)); //fast, but will nullify u and v
Matrix w = MyAlgorithm(u, move(v)); //and you can even do this if you need one but not the other

Related

I am pretty confused on the mental model for move semantics in C++ [duplicate]

What is it?
What does it do?
When should it be used?
Good links are appreciated.
1. "What is it?"
While std::move() is technically a function - I would say it isn't really a function. It's sort of a converter between ways the compiler considers an expression's value.
2. "What does it do?"
The first thing to note is that std::move() doesn't actually move anything. It changes an expression from being an lvalue (such as a named variable) to being an xvalue. An xvalue tells the compiler:
You can plunder me, move anything I'm holding and use it elsewhere (since I'm going to be destroyed soon anyway)".
in other words, when you use std::move(x), you're allowing the compiler to cannibalize x. Thus if x has, say, its own buffer in memory - after std::move()ing the compiler can have another object own it instead.
You can also move from a prvalue (such as a temporary you're passing around), but this is rarely useful.
3. "When should it be used?"
Another way to ask this question is "What would I cannibalize an existing object's resources for?" well, if you're writing application code, you would probably not be messing around a lot with temporary objects created by the compiler. So mainly you would do this in places like constructors, operator methods, standard-library-algorithm-like functions etc. where objects get created and destroyed automagically a lot. Of course, that's just a rule of thumb.
A typical use is 'moving' resources from one object to another instead of copying. #Guillaume links to this page which has a straightforward short example: swapping two objects with less copying.
template <class T>
swap(T& a, T& b) {
T tmp(a); // we now have two copies of a
a = b; // we now have two copies of b (+ discarded a copy of a)
b = tmp; // we now have two copies of tmp (+ discarded a copy of b)
}
using move allows you to swap the resources instead of copying them around:
template <class T>
swap(T& a, T& b) {
T tmp(std::move(a));
a = std::move(b);
b = std::move(tmp);
}
Think of what happens when T is, say, vector<int> of size n. In the first version you read and write 3*n elements, in the second version you basically read and write just the 3 pointers to the vectors' buffers, plus the 3 buffers' sizes. Of course, class T needs to know how to do the moving; your class should have a move-assignment operator and a move-constructor for class T for this to work.
Wikipedia Page on C++11 R-value references and move constructors
In C++11, in addition to copy constructors, objects can have move constructors.
(And in addition to copy assignment operators, they have move assignment operators.)
The move constructor is used instead of the copy constructor, if the object has type "rvalue-reference" (Type &&).
std::move() is a cast that produces an rvalue-reference to an object, to enable moving from it.
It's a new C++ way to avoid copies. For example, using a move constructor, a std::vector could just copy its internal pointer to data to the new object, leaving the moved object in an moved from state, therefore not copying all the data. This would be C++-valid.
Try googling for move semantics, rvalue, perfect forwarding.
You can use move when you need to "transfer" the content of an object somewhere else, without doing a copy (i.e. the content is not duplicated, that's why it could be used on some non-copyable objects, like a unique_ptr). It's also possible for an object to take the content of a temporary object without doing a copy (and save a lot of time), with std::move.
This link really helped me out :
http://thbecker.net/articles/rvalue_references/section_01.html
I'm sorry if my answer is coming too late, but I was also looking for a good link for the std::move, and I found the links above a little bit "austere".
This puts the emphasis on r-value reference, in which context you should use them, and I think it's more detailed, that's why I wanted to share this link here.
Q: What is std::move?
A: std::move() is a function from the C++ Standard Library for casting to a rvalue reference.
Simplisticly std::move(t) is equivalent to:
static_cast<T&&>(t);
An rvalue is a temporary that does not persist beyond the expression that defines it, such as an intermediate function result which is never stored in a variable.
int a = 3; // 3 is a rvalue, does not exist after expression is evaluated
int b = a; // a is a lvalue, keeps existing after expression is evaluated
An implementation for std::move() is given in N2027: "A Brief Introduction to Rvalue References" as follows:
template <class T>
typename remove_reference<T>::type&&
std::move(T&& a)
{
return a;
}
As you can see, std::move returns T&& no matter if called with a value (T), reference type (T&), or rvalue reference (T&&).
Q: What does it do?
A: As a cast, it does not do anything during runtime. It is only relevant at compile time to tell the compiler that you would like to continue considering the reference as an rvalue.
foo(3 * 5); // obviously, you are calling foo with a temporary (rvalue)
int a = 3 * 5;
foo(a); // how to tell the compiler to treat `a` as an rvalue?
foo(std::move(a)); // will call `foo(int&& a)` rather than `foo(int a)` or `foo(int& a)`
What it does not do:
Make a copy of the argument
Call the copy constructor
Change the argument object
Q: When should it be used?
A: You should use std::move if you want to call functions that support move semantics with an argument which is not an rvalue (temporary expression).
This begs the following follow-up questions for me:
What is move semantics? Move semantics in contrast to copy semantics is a programming technique in which the members of an object are initialized by 'taking over' instead of copying another object's members. Such 'take over' makes only sense with pointers and resource handles, which can be cheaply transferred by copying the pointer or integer handle rather than the underlying data.
What kind of classes and objects support move semantics? It is up to you as a developer to implement move semantics in your own classes if these would benefit from transferring their members instead of copying them. Once you implement move semantics, you will directly benefit from work from many library programmers who have added support for handling classes with move semantics efficiently.
Why can't the compiler figure it out on its own? The compiler cannot just call another overload of a function unless you say so. You must help the compiler choose whether the regular or move version of the function should be called.
In which situations would I want to tell the compiler that it should treat a variable as an rvalue? This will most likely happen in template or library functions, where you know that an intermediate result could be salvaged (rather than allocating a new instance).
std::move itself doesn't really do much. I thought that it called the moved constructor for an object, but it really just performs a type cast (casting an lvalue variable to an rvalue so that the said variable can be passed as an argument to a move constructor or assignment operator).
So std::move is just used as a precursor to using move semantics. Move semantics is essentially an efficient way for dealing with temporary objects.
Consider Object A = B + (C + (D + (E + F)));
This is nice looking code, but E + F produces a temporary object. Then D + temp produces another temporary object and so on. In each normal "+" operator of a class, deep copies occur.
For example
Object Object::operator+ (const Object& rhs) {
Object temp (*this);
// logic for adding
return temp;
}
The creation of the temporary object in this function is useless - these temporary objects will be deleted at the end of the line anyway as they go out of scope.
We can rather use move semantics to "plunder" the temporary objects and do something like
Object& Object::operator+ (Object&& rhs) {
// logic to modify rhs directly
return rhs;
}
This avoids needless deep copies being made. With reference to the example, the only part where deep copying occurs is now E + F. The rest uses move semantics. The move constructor or assignment operator also needs to be implemented to assign the result to A.
"What is it?" and "What does it do?" has been explained above.
I will give a example of "when it should be used".
For example, we have a class with lots of resource like big array in it.
class ResHeavy{ // ResHeavy means heavy resource
public:
ResHeavy(int len=10):_upInt(new int[len]),_len(len){
cout<<"default ctor"<<endl;
}
ResHeavy(const ResHeavy& rhs):_upInt(new int[rhs._len]),_len(rhs._len){
cout<<"copy ctor"<<endl;
}
ResHeavy& operator=(const ResHeavy& rhs){
_upInt.reset(new int[rhs._len]);
_len = rhs._len;
cout<<"operator= ctor"<<endl;
}
ResHeavy(ResHeavy&& rhs){
_upInt = std::move(rhs._upInt);
_len = rhs._len;
rhs._len = 0;
cout<<"move ctor"<<endl;
}
// check array valid
bool is_up_valid(){
return _upInt != nullptr;
}
private:
std::unique_ptr<int[]> _upInt; // heavy array resource
int _len; // length of int array
};
Test code:
void test_std_move2(){
ResHeavy rh; // only one int[]
// operator rh
// after some operator of rh, it becomes no-use
// transform it to other object
ResHeavy rh2 = std::move(rh); // rh becomes invalid
// show rh, rh2 it valid
if(rh.is_up_valid())
cout<<"rh valid"<<endl;
else
cout<<"rh invalid"<<endl;
if(rh2.is_up_valid())
cout<<"rh2 valid"<<endl;
else
cout<<"rh2 invalid"<<endl;
// new ResHeavy object, created by copy ctor
ResHeavy rh3(rh2); // two copy of int[]
if(rh3.is_up_valid())
cout<<"rh3 valid"<<endl;
else
cout<<"rh3 invalid"<<endl;
}
output as below:
default ctor
move ctor
rh invalid
rh2 valid
copy ctor
rh3 valid
We can see that std::move with move constructor makes transform resource easily.
Where else is std::move useful?
std::move can also be useful when sorting an array of elements. Many sorting algorithms (such as selection sort and bubble sort) work by swapping pairs of elements. Previously, we’ve had to resort to copy-semantics to do the swapping. Now we can use move semantics, which is more efficient.
It can also be useful if we want to move the contents managed by one smart pointer to another.
Cited:
https://www.learncpp.com/cpp-tutorial/15-4-stdmove/
std::move itself does nothing rather than a static_cast. According to cppreference.com
It is exactly equivalent to a static_cast to an rvalue reference type.
Thus, it depends on the type of the variable you assign to after the move, if the type has constructors or assign operators that takes a rvalue parameter, it may or may not steal the content of the original variable, so, it may leave the original variable to be in an unspecified state:
Unless otherwise specified, all standard library objects that have been moved from being placed in a valid but unspecified state.
Because there is no special move constructor or move assign operator for built-in literal types such as integers and raw pointers, so, it will be just a simple copy for these types.
Here is a full example, using std::move for a (simple) custom vector
Expected output:
c: [10][11]
copy ctor called
copy of c: [10][11]
move ctor called
moved c: [10][11]
Compile as:
g++ -std=c++2a -O2 -Wall -pedantic foo.cpp
Code:
#include <iostream>
#include <algorithm>
template<class T> class MyVector {
private:
T *data;
size_t maxlen;
size_t currlen;
public:
MyVector<T> () : data (nullptr), maxlen(0), currlen(0) { }
MyVector<T> (int maxlen) : data (new T [maxlen]), maxlen(maxlen), currlen(0) { }
MyVector<T> (const MyVector& o) {
std::cout << "copy ctor called" << std::endl;
data = new T [o.maxlen];
maxlen = o.maxlen;
currlen = o.currlen;
std::copy(o.data, o.data + o.maxlen, data);
}
MyVector<T> (const MyVector<T>&& o) {
std::cout << "move ctor called" << std::endl;
data = o.data;
maxlen = o.maxlen;
currlen = o.currlen;
}
void push_back (const T& i) {
if (currlen >= maxlen) {
maxlen *= 2;
auto newdata = new T [maxlen];
std::copy(data, data + currlen, newdata);
if (data) {
delete[] data;
}
data = newdata;
}
data[currlen++] = i;
}
friend std::ostream& operator<<(std::ostream &os, const MyVector<T>& o) {
auto s = o.data;
auto e = o.data + o.currlen;;
while (s < e) {
os << "[" << *s << "]";
s++;
}
return os;
}
};
int main() {
auto c = new MyVector<int>(1);
c->push_back(10);
c->push_back(11);
std::cout << "c: " << *c << std::endl;
auto d = *c;
std::cout << "copy of c: " << d << std::endl;
auto e = std::move(*c);
delete c;
std::cout << "moved c: " << e << std::endl;
}
std::move simply casts a variable to an rvalue reference. This rvalue reference is notated with &&. Let's say you have a class Foo and you instantiate an object like this
Foo foo = Foo();
If you then write
Foo foo2 = std::move(foo);
that's the same thing as If I wrote
Foo foo2 = (Foo&&) foo;
std::move replaces this cast to an rvalue reference.
The reason why you would want to write either of the previous 2 lines of code
is that if you write
Foo foo2 = foo;
The copy constructor will be called.
Let's say Foo instances have a pointer to some data on the heap which they own.
In Foo's destructor that data on the heap gets deleted.
If you want to distinghuish between copying the data from the heap and taking ownership of that data, you can write a constructor which takes in const Foo& and that constructor can perform the deep copy. Then you can write a constructor which takes in an rvalue reference (Foo&&) and this constructor can simply rewire the pointers.
This constructor which takes in Foo&& will be called when you write
Foo foo2 = std::move(foo);
and when you write
Foo foo2 = (Foo&&) foo;

Difference between std::move and assignment in C++ [duplicate]

What is it?
What does it do?
When should it be used?
Good links are appreciated.
1. "What is it?"
While std::move() is technically a function - I would say it isn't really a function. It's sort of a converter between ways the compiler considers an expression's value.
2. "What does it do?"
The first thing to note is that std::move() doesn't actually move anything. It changes an expression from being an lvalue (such as a named variable) to being an xvalue. An xvalue tells the compiler:
You can plunder me, move anything I'm holding and use it elsewhere (since I'm going to be destroyed soon anyway)".
in other words, when you use std::move(x), you're allowing the compiler to cannibalize x. Thus if x has, say, its own buffer in memory - after std::move()ing the compiler can have another object own it instead.
You can also move from a prvalue (such as a temporary you're passing around), but this is rarely useful.
3. "When should it be used?"
Another way to ask this question is "What would I cannibalize an existing object's resources for?" well, if you're writing application code, you would probably not be messing around a lot with temporary objects created by the compiler. So mainly you would do this in places like constructors, operator methods, standard-library-algorithm-like functions etc. where objects get created and destroyed automagically a lot. Of course, that's just a rule of thumb.
A typical use is 'moving' resources from one object to another instead of copying. #Guillaume links to this page which has a straightforward short example: swapping two objects with less copying.
template <class T>
swap(T& a, T& b) {
T tmp(a); // we now have two copies of a
a = b; // we now have two copies of b (+ discarded a copy of a)
b = tmp; // we now have two copies of tmp (+ discarded a copy of b)
}
using move allows you to swap the resources instead of copying them around:
template <class T>
swap(T& a, T& b) {
T tmp(std::move(a));
a = std::move(b);
b = std::move(tmp);
}
Think of what happens when T is, say, vector<int> of size n. In the first version you read and write 3*n elements, in the second version you basically read and write just the 3 pointers to the vectors' buffers, plus the 3 buffers' sizes. Of course, class T needs to know how to do the moving; your class should have a move-assignment operator and a move-constructor for class T for this to work.
Wikipedia Page on C++11 R-value references and move constructors
In C++11, in addition to copy constructors, objects can have move constructors.
(And in addition to copy assignment operators, they have move assignment operators.)
The move constructor is used instead of the copy constructor, if the object has type "rvalue-reference" (Type &&).
std::move() is a cast that produces an rvalue-reference to an object, to enable moving from it.
It's a new C++ way to avoid copies. For example, using a move constructor, a std::vector could just copy its internal pointer to data to the new object, leaving the moved object in an moved from state, therefore not copying all the data. This would be C++-valid.
Try googling for move semantics, rvalue, perfect forwarding.
You can use move when you need to "transfer" the content of an object somewhere else, without doing a copy (i.e. the content is not duplicated, that's why it could be used on some non-copyable objects, like a unique_ptr). It's also possible for an object to take the content of a temporary object without doing a copy (and save a lot of time), with std::move.
This link really helped me out :
http://thbecker.net/articles/rvalue_references/section_01.html
I'm sorry if my answer is coming too late, but I was also looking for a good link for the std::move, and I found the links above a little bit "austere".
This puts the emphasis on r-value reference, in which context you should use them, and I think it's more detailed, that's why I wanted to share this link here.
Q: What is std::move?
A: std::move() is a function from the C++ Standard Library for casting to a rvalue reference.
Simplisticly std::move(t) is equivalent to:
static_cast<T&&>(t);
An rvalue is a temporary that does not persist beyond the expression that defines it, such as an intermediate function result which is never stored in a variable.
int a = 3; // 3 is a rvalue, does not exist after expression is evaluated
int b = a; // a is a lvalue, keeps existing after expression is evaluated
An implementation for std::move() is given in N2027: "A Brief Introduction to Rvalue References" as follows:
template <class T>
typename remove_reference<T>::type&&
std::move(T&& a)
{
return a;
}
As you can see, std::move returns T&& no matter if called with a value (T), reference type (T&), or rvalue reference (T&&).
Q: What does it do?
A: As a cast, it does not do anything during runtime. It is only relevant at compile time to tell the compiler that you would like to continue considering the reference as an rvalue.
foo(3 * 5); // obviously, you are calling foo with a temporary (rvalue)
int a = 3 * 5;
foo(a); // how to tell the compiler to treat `a` as an rvalue?
foo(std::move(a)); // will call `foo(int&& a)` rather than `foo(int a)` or `foo(int& a)`
What it does not do:
Make a copy of the argument
Call the copy constructor
Change the argument object
Q: When should it be used?
A: You should use std::move if you want to call functions that support move semantics with an argument which is not an rvalue (temporary expression).
This begs the following follow-up questions for me:
What is move semantics? Move semantics in contrast to copy semantics is a programming technique in which the members of an object are initialized by 'taking over' instead of copying another object's members. Such 'take over' makes only sense with pointers and resource handles, which can be cheaply transferred by copying the pointer or integer handle rather than the underlying data.
What kind of classes and objects support move semantics? It is up to you as a developer to implement move semantics in your own classes if these would benefit from transferring their members instead of copying them. Once you implement move semantics, you will directly benefit from work from many library programmers who have added support for handling classes with move semantics efficiently.
Why can't the compiler figure it out on its own? The compiler cannot just call another overload of a function unless you say so. You must help the compiler choose whether the regular or move version of the function should be called.
In which situations would I want to tell the compiler that it should treat a variable as an rvalue? This will most likely happen in template or library functions, where you know that an intermediate result could be salvaged (rather than allocating a new instance).
std::move itself doesn't really do much. I thought that it called the moved constructor for an object, but it really just performs a type cast (casting an lvalue variable to an rvalue so that the said variable can be passed as an argument to a move constructor or assignment operator).
So std::move is just used as a precursor to using move semantics. Move semantics is essentially an efficient way for dealing with temporary objects.
Consider Object A = B + (C + (D + (E + F)));
This is nice looking code, but E + F produces a temporary object. Then D + temp produces another temporary object and so on. In each normal "+" operator of a class, deep copies occur.
For example
Object Object::operator+ (const Object& rhs) {
Object temp (*this);
// logic for adding
return temp;
}
The creation of the temporary object in this function is useless - these temporary objects will be deleted at the end of the line anyway as they go out of scope.
We can rather use move semantics to "plunder" the temporary objects and do something like
Object& Object::operator+ (Object&& rhs) {
// logic to modify rhs directly
return rhs;
}
This avoids needless deep copies being made. With reference to the example, the only part where deep copying occurs is now E + F. The rest uses move semantics. The move constructor or assignment operator also needs to be implemented to assign the result to A.
"What is it?" and "What does it do?" has been explained above.
I will give a example of "when it should be used".
For example, we have a class with lots of resource like big array in it.
class ResHeavy{ // ResHeavy means heavy resource
public:
ResHeavy(int len=10):_upInt(new int[len]),_len(len){
cout<<"default ctor"<<endl;
}
ResHeavy(const ResHeavy& rhs):_upInt(new int[rhs._len]),_len(rhs._len){
cout<<"copy ctor"<<endl;
}
ResHeavy& operator=(const ResHeavy& rhs){
_upInt.reset(new int[rhs._len]);
_len = rhs._len;
cout<<"operator= ctor"<<endl;
}
ResHeavy(ResHeavy&& rhs){
_upInt = std::move(rhs._upInt);
_len = rhs._len;
rhs._len = 0;
cout<<"move ctor"<<endl;
}
// check array valid
bool is_up_valid(){
return _upInt != nullptr;
}
private:
std::unique_ptr<int[]> _upInt; // heavy array resource
int _len; // length of int array
};
Test code:
void test_std_move2(){
ResHeavy rh; // only one int[]
// operator rh
// after some operator of rh, it becomes no-use
// transform it to other object
ResHeavy rh2 = std::move(rh); // rh becomes invalid
// show rh, rh2 it valid
if(rh.is_up_valid())
cout<<"rh valid"<<endl;
else
cout<<"rh invalid"<<endl;
if(rh2.is_up_valid())
cout<<"rh2 valid"<<endl;
else
cout<<"rh2 invalid"<<endl;
// new ResHeavy object, created by copy ctor
ResHeavy rh3(rh2); // two copy of int[]
if(rh3.is_up_valid())
cout<<"rh3 valid"<<endl;
else
cout<<"rh3 invalid"<<endl;
}
output as below:
default ctor
move ctor
rh invalid
rh2 valid
copy ctor
rh3 valid
We can see that std::move with move constructor makes transform resource easily.
Where else is std::move useful?
std::move can also be useful when sorting an array of elements. Many sorting algorithms (such as selection sort and bubble sort) work by swapping pairs of elements. Previously, we’ve had to resort to copy-semantics to do the swapping. Now we can use move semantics, which is more efficient.
It can also be useful if we want to move the contents managed by one smart pointer to another.
Cited:
https://www.learncpp.com/cpp-tutorial/15-4-stdmove/
std::move itself does nothing rather than a static_cast. According to cppreference.com
It is exactly equivalent to a static_cast to an rvalue reference type.
Thus, it depends on the type of the variable you assign to after the move, if the type has constructors or assign operators that takes a rvalue parameter, it may or may not steal the content of the original variable, so, it may leave the original variable to be in an unspecified state:
Unless otherwise specified, all standard library objects that have been moved from being placed in a valid but unspecified state.
Because there is no special move constructor or move assign operator for built-in literal types such as integers and raw pointers, so, it will be just a simple copy for these types.
Here is a full example, using std::move for a (simple) custom vector
Expected output:
c: [10][11]
copy ctor called
copy of c: [10][11]
move ctor called
moved c: [10][11]
Compile as:
g++ -std=c++2a -O2 -Wall -pedantic foo.cpp
Code:
#include <iostream>
#include <algorithm>
template<class T> class MyVector {
private:
T *data;
size_t maxlen;
size_t currlen;
public:
MyVector<T> () : data (nullptr), maxlen(0), currlen(0) { }
MyVector<T> (int maxlen) : data (new T [maxlen]), maxlen(maxlen), currlen(0) { }
MyVector<T> (const MyVector& o) {
std::cout << "copy ctor called" << std::endl;
data = new T [o.maxlen];
maxlen = o.maxlen;
currlen = o.currlen;
std::copy(o.data, o.data + o.maxlen, data);
}
MyVector<T> (const MyVector<T>&& o) {
std::cout << "move ctor called" << std::endl;
data = o.data;
maxlen = o.maxlen;
currlen = o.currlen;
}
void push_back (const T& i) {
if (currlen >= maxlen) {
maxlen *= 2;
auto newdata = new T [maxlen];
std::copy(data, data + currlen, newdata);
if (data) {
delete[] data;
}
data = newdata;
}
data[currlen++] = i;
}
friend std::ostream& operator<<(std::ostream &os, const MyVector<T>& o) {
auto s = o.data;
auto e = o.data + o.currlen;;
while (s < e) {
os << "[" << *s << "]";
s++;
}
return os;
}
};
int main() {
auto c = new MyVector<int>(1);
c->push_back(10);
c->push_back(11);
std::cout << "c: " << *c << std::endl;
auto d = *c;
std::cout << "copy of c: " << d << std::endl;
auto e = std::move(*c);
delete c;
std::cout << "moved c: " << e << std::endl;
}
std::move simply casts a variable to an rvalue reference. This rvalue reference is notated with &&. Let's say you have a class Foo and you instantiate an object like this
Foo foo = Foo();
If you then write
Foo foo2 = std::move(foo);
that's the same thing as If I wrote
Foo foo2 = (Foo&&) foo;
std::move replaces this cast to an rvalue reference.
The reason why you would want to write either of the previous 2 lines of code
is that if you write
Foo foo2 = foo;
The copy constructor will be called.
Let's say Foo instances have a pointer to some data on the heap which they own.
In Foo's destructor that data on the heap gets deleted.
If you want to distinghuish between copying the data from the heap and taking ownership of that data, you can write a constructor which takes in const Foo& and that constructor can perform the deep copy. Then you can write a constructor which takes in an rvalue reference (Foo&&) and this constructor can simply rewire the pointers.
This constructor which takes in Foo&& will be called when you write
Foo foo2 = std::move(foo);
and when you write
Foo foo2 = (Foo&&) foo;

When move semantics required on practice (in case there is copy elision) [duplicate]

As far as I understand one of the purposes of adding move semantics is to optimize code by calling special constructor for copying "temporary" objects. For example, in this answer we see that it can be used to optimize such string a = x + y stuff. Because x+y is an rvalue expression, instead of deep copying we can copy only the pointer to the string and the size of the string. But as we know, modern compilers support return value optimization, so without using move semantics our code will not call the copy constructor at all.
To prove it I write this code:
#include <iostream>
struct stuff
{
int x;
stuff(int x_):x(x_){}
stuff(const stuff & g):x(g.x)
{
std::cout<<"copy"<<std::endl;
}
};
stuff operator+(const stuff& lhs,const stuff& rhs)
{
stuff g(lhs.x+rhs.x);
return g;
}
int main()
{
stuff a(5),b(7);
stuff c = a+b;
}
And after executing it in VC++2010 and g++ in optimize mode I'm getting empty output.
What kind of optimization is it, if without it my code still works faster? Could you explain what I'm understanding wrong?
Move semantics should not be thought as an optimization device, even if they can be used as such.
If you are going to want copies of objects (either function parameters or return values), then RVO and copy elision will do the job when they can. Move semantics can help, but are more powerful than that.
Move semantics are handy when you want to do something different whether the passed object is a temporary (it then binds to a rvalue reference) or a "standard" object with a name (a so called const lvalue). If you want for instance to steal the resources of a temporary object, then you want move semantics (example: you can steal the contents a std::unique_ptr points to).
Move semantics allow you to return non copyable objects from functions, which is not possible with the current standard. Also, non copyable objects can be put inside other objects, and those objects will automatically be movable if the contained objects are.
Non copyable objects are great, since they don't force you to implement an error-prone copy constructor. A lot of the time, copy semantics do not really make sense, but move semantics do (think about it).
This also enables you to use movable std::vector<T> classes even if T is non copyable. The std::unique_ptr class template is also a great tool when dealing with non copyable objects (eg. polymorphic objects).
After some digging I find this excellent example of optimization with rvalue references inStroustrup's FAQ .
Yes, swap function:
template<class T>
void swap(T& a, T& b) // "perfect swap" (almost)
{
T tmp = move(a); // could invalidate a
a = move(b); // could invalidate b
b = move(tmp); // could invalidate tmp
}
This will generate optimized code for any kind of types (assuming, that it have move constructor).
Edit: Also RVO can't optimize something like this(at least on my compiler):
stuff func(const stuff& st)
{
if(st.x>0)
{
stuff ret(2*st.x);
return ret;
}
else
{
stuff ret2(-2*st.x);
return ret2;
}
}
This function always calls copy constructor (checked with VC++). And if our class can be moved faster, than with move constructor we will have optimization.
Imagine your stuff was a class with heap allocated memory like a string, and that it had the notion of capacity. Give it a operator+= that will grow the capacity geometrically. In C++03 this might look like:
#include <iostream>
#include <algorithm>
struct stuff
{
int size;
int cap;
stuff(int size_):size(size_)
{
cap = size;
if (cap > 0)
std::cout <<"allocating " << cap <<std::endl;
}
stuff(const stuff & g):size(g.size), cap(g.cap)
{
if (cap > 0)
std::cout <<"allocating " << cap <<std::endl;
}
~stuff()
{
if (cap > 0)
std::cout << "deallocating " << cap << '\n';
}
stuff& operator+=(const stuff& y)
{
if (cap < size+y.size)
{
if (cap > 0)
std::cout << "deallocating " << cap << '\n';
cap = std::max(2*cap, size+y.size);
std::cout <<"allocating " << cap <<std::endl;
}
size += y.size;
return *this;
}
};
stuff operator+(const stuff& lhs,const stuff& rhs)
{
stuff g(lhs.size + rhs.size);
return g;
}
Also imagine you want to add more than just two stuff's at a time:
int main()
{
stuff a(11),b(9),c(7),d(5);
std::cout << "start addition\n\n";
stuff e = a+b+c+d;
std::cout << "\nend addition\n";
}
For me this prints out:
allocating 11
allocating 9
allocating 7
allocating 5
start addition
allocating 20
allocating 27
allocating 32
deallocating 27
deallocating 20
end addition
deallocating 32
deallocating 5
deallocating 7
deallocating 9
deallocating 11
I count 3 allocations and 2 deallocations to compute:
stuff e = a+b+c+d;
Now add move semantics:
stuff(stuff&& g):size(g.size), cap(g.cap)
{
g.cap = 0;
g.size = 0;
}
...
stuff operator+(stuff&& lhs,const stuff& rhs)
{
return std::move(lhs += rhs);
}
Running again I get:
allocating 11
allocating 9
allocating 7
allocating 5
start addition
allocating 20
deallocating 20
allocating 40
end addition
deallocating 40
deallocating 5
deallocating 7
deallocating 9
deallocating 11
I'm now down to 2 allocations and 1 deallocations. That translates to faster code.
There are many places some of which are mentioned in other answers.
One big one is that when resizing a std::vector it will move move-aware objects from the old memory location to the new one rather than copy and destroy the original.
Additionally rvalue references allow the concept of movable types, this is a semantic difference and not just an optimization. unique_ptr wasn't possible in C++03 which is why we had the abomination of auto_ptr.
Just because this particular case is already covered by an existing optimization does not mean that other cases don't exist where r-value references are helpful.
Move construction allows optimization even when the temporary is returned from a function which cannot be inlined (perhaps it's a virtual call, or through a function pointer).
Your posted example only takes const lvalue references and so explicitly cannot have move semantics applied to it, as there is not a single rvalue reference in there. How can move semantics make your code faster when you implemented a type without rvalue references?
In addition, your code is already covered by RVO and NRVO. Move semantics apply to far, far more situations than those two do.
This line calls the first constructor.
stuff a(5),b(7);
Plus operator is called using explicit common lvalue references.
stuff c = a + b;
Inside operator overload method, you have no copy constructor called.
Again, the first constructor is called only.
stuff g(lhs.x+rhs.x);
assigment is made with RVO, so no copy is need. NO copy from returned object to 'c' is need.
stuff c = a+b;
Due no std::cout reference, compiler take care about your c value is never used. Then, whole program is stripped out, resulting in a empty program.
Another good example I can think of. Imagine that you're implementing a matrix library and write an algorithm which takes two matrices and outputs another one:
Matrix MyAlgorithm(Matrix U, Matrix V)
{
Transform(U); //doesn't matter what this actually does, but it modifies U
Transform(V);
return U*V;
}
Note that you can't pass U and V by const reference, because the algorithm tweaks them. You could theoretically pass them by reference, but this would look gross and leave U and V in some intermediate state (since you call Transform(U)), which may not make any sense to the caller, or just not make any mathematical sense at all, since it's just one of the internal algorithm transformations. The code looks much cleaner if you just pass them by value and use move semantics if you are not going to use U and V after calling this function:
Matrix u, v;
...
Matrix w = MyAlgorithm(u, v); //slow, but will preserve u and v
Matrix w = MyAlgorithm(move(u), move(v)); //fast, but will nullify u and v
Matrix w = MyAlgorithm(u, move(v)); //and you can even do this if you need one but not the other

Is pass by value that much faster?

I've heard that you should always prefer "pass by value" in C++11 because of the introduction of move semantics. I wanted to see what the hype was all about and constructed a test case. First my class:
struct MyClass {
MyClass() { }
MyClass(const MyClass&) { std::cout << "Copy construct" << std::endl; }
MyClass(MyClass&&) { std::cout << "Move construct" << std::endl; }
~MyClass() { }
};
And the test harness:
class Test
{
public:
void pass_by_lvalue_ref(const MyClass& myClass)
{
_MyClass.push_back(myClass);
}
void pass_by_rvalue_ref(MyClass&& myClass)
{
_MyClass.push_back(std::move(myClass));
}
void pass_by_value(MyClass myClass)
{
_MyClass.push_back(std::move(myClass));
}
private:
std::vector<MyClass> _MyClass;
};
Presumably, pass_by_value should outperform pass_by_lvalue_ref and pass_by_rvalue_ref (together, not separately).
int main()
{
MyClass myClass;
Test Test;
std::cout << "--lvalue_ref--\n";
Test.pass_by_lvalue_ref(myClass);
std::cout << "--rvalue_ref--\n";
Test.pass_by_rvalue_ref(MyClass{});
std::cout << "--value - lvalue--\n";
Test.pass_by_value(myClass);
std::cout << "--value - rvalue--\n";
Test.pass_by_value(MyClass{});
}
This is my output on GCC 4.9.2 with -O2:
--lvalue_ref--
Copy construct
--rvalue_ref--
Move construct
Copy construct
--value - lvalue--
Copy construct
Move construct
Copy construct
Copy construct
--value - rvalue--
Move construct
As you can see, the non-pass_by_value functions requires a total of 2 copy constructs and 1 move construct. The pass_by_value function requires a total of 3 copy constructs and 2 move constructs. It looks like that, as expected, the object is going to be copied anyway, so why does everyone say pass by value?
First, your reporting is entirely flawed. Each of your functions pushes back to the same vector. When that vector runs out of capacity (which depends upon how many items you've inserted so far), it is going to trigger a re-allocation which will require more moves and/or copies than an insertion which doesn't trigger an allocation.
Second, std::vector::push_back has a strong exception safety guarantee. So if your move constructor is not noexcept, it will not use it (unless the class is non-copyable). It will use the copy constructor instead.
Third,
I've heard that you should always prefer "pass by value" in C++11
because of the introduction of move semantics.
I'm pretty sure you didn't hear that from any reputable source. Or are actually inappropriately paraphrasing what was actually said. But I don't have the source of the quote. What is usually advised is actually that if you are going to copy your arguments in your function anyway, don't. Just do it in the parameter list (via pass by value). This will allow your function to move r-value arguments straight to their destination. When you pass l-values, they will be copied, but you were going to do that anyway.
If you are going to make an internal copy, then passing by value will do exactly one move construct more than the pair of overloads (pass by rvalue ref)+(pass by const lvalue ref).
If move construct is cheap, this is a small amount of runtime overhead in exchange for less compile time and code maintenance overhead.
The idiom is "Want speed? Making a copy anyhow? Pass by value, instead of by const lvalue reference." in reality.
Finally, your benchmark is flawed as you failed to reserve(enough) before your push backs. Reallocation can cause extra operations. Oh, and make your move constructor noexcept, as conforming libraries will prefer a copy to a move if move can throw in many situations.

What is std::move(), and when should it be used?

What is it?
What does it do?
When should it be used?
Good links are appreciated.
1. "What is it?"
While std::move() is technically a function - I would say it isn't really a function. It's sort of a converter between ways the compiler considers an expression's value.
2. "What does it do?"
The first thing to note is that std::move() doesn't actually move anything. It changes an expression from being an lvalue (such as a named variable) to being an xvalue. An xvalue tells the compiler:
You can plunder me, move anything I'm holding and use it elsewhere (since I'm going to be destroyed soon anyway)".
in other words, when you use std::move(x), you're allowing the compiler to cannibalize x. Thus if x has, say, its own buffer in memory - after std::move()ing the compiler can have another object own it instead.
You can also move from a prvalue (such as a temporary you're passing around), but this is rarely useful.
3. "When should it be used?"
Another way to ask this question is "What would I cannibalize an existing object's resources for?" well, if you're writing application code, you would probably not be messing around a lot with temporary objects created by the compiler. So mainly you would do this in places like constructors, operator methods, standard-library-algorithm-like functions etc. where objects get created and destroyed automagically a lot. Of course, that's just a rule of thumb.
A typical use is 'moving' resources from one object to another instead of copying. #Guillaume links to this page which has a straightforward short example: swapping two objects with less copying.
template <class T>
swap(T& a, T& b) {
T tmp(a); // we now have two copies of a
a = b; // we now have two copies of b (+ discarded a copy of a)
b = tmp; // we now have two copies of tmp (+ discarded a copy of b)
}
using move allows you to swap the resources instead of copying them around:
template <class T>
swap(T& a, T& b) {
T tmp(std::move(a));
a = std::move(b);
b = std::move(tmp);
}
Think of what happens when T is, say, vector<int> of size n. In the first version you read and write 3*n elements, in the second version you basically read and write just the 3 pointers to the vectors' buffers, plus the 3 buffers' sizes. Of course, class T needs to know how to do the moving; your class should have a move-assignment operator and a move-constructor for class T for this to work.
Wikipedia Page on C++11 R-value references and move constructors
In C++11, in addition to copy constructors, objects can have move constructors.
(And in addition to copy assignment operators, they have move assignment operators.)
The move constructor is used instead of the copy constructor, if the object has type "rvalue-reference" (Type &&).
std::move() is a cast that produces an rvalue-reference to an object, to enable moving from it.
It's a new C++ way to avoid copies. For example, using a move constructor, a std::vector could just copy its internal pointer to data to the new object, leaving the moved object in an moved from state, therefore not copying all the data. This would be C++-valid.
Try googling for move semantics, rvalue, perfect forwarding.
You can use move when you need to "transfer" the content of an object somewhere else, without doing a copy (i.e. the content is not duplicated, that's why it could be used on some non-copyable objects, like a unique_ptr). It's also possible for an object to take the content of a temporary object without doing a copy (and save a lot of time), with std::move.
This link really helped me out :
http://thbecker.net/articles/rvalue_references/section_01.html
I'm sorry if my answer is coming too late, but I was also looking for a good link for the std::move, and I found the links above a little bit "austere".
This puts the emphasis on r-value reference, in which context you should use them, and I think it's more detailed, that's why I wanted to share this link here.
Q: What is std::move?
A: std::move() is a function from the C++ Standard Library for casting to a rvalue reference.
Simplisticly std::move(t) is equivalent to:
static_cast<T&&>(t);
An rvalue is a temporary that does not persist beyond the expression that defines it, such as an intermediate function result which is never stored in a variable.
int a = 3; // 3 is a rvalue, does not exist after expression is evaluated
int b = a; // a is a lvalue, keeps existing after expression is evaluated
An implementation for std::move() is given in N2027: "A Brief Introduction to Rvalue References" as follows:
template <class T>
typename remove_reference<T>::type&&
std::move(T&& a)
{
return a;
}
As you can see, std::move returns T&& no matter if called with a value (T), reference type (T&), or rvalue reference (T&&).
Q: What does it do?
A: As a cast, it does not do anything during runtime. It is only relevant at compile time to tell the compiler that you would like to continue considering the reference as an rvalue.
foo(3 * 5); // obviously, you are calling foo with a temporary (rvalue)
int a = 3 * 5;
foo(a); // how to tell the compiler to treat `a` as an rvalue?
foo(std::move(a)); // will call `foo(int&& a)` rather than `foo(int a)` or `foo(int& a)`
What it does not do:
Make a copy of the argument
Call the copy constructor
Change the argument object
Q: When should it be used?
A: You should use std::move if you want to call functions that support move semantics with an argument which is not an rvalue (temporary expression).
This begs the following follow-up questions for me:
What is move semantics? Move semantics in contrast to copy semantics is a programming technique in which the members of an object are initialized by 'taking over' instead of copying another object's members. Such 'take over' makes only sense with pointers and resource handles, which can be cheaply transferred by copying the pointer or integer handle rather than the underlying data.
What kind of classes and objects support move semantics? It is up to you as a developer to implement move semantics in your own classes if these would benefit from transferring their members instead of copying them. Once you implement move semantics, you will directly benefit from work from many library programmers who have added support for handling classes with move semantics efficiently.
Why can't the compiler figure it out on its own? The compiler cannot just call another overload of a function unless you say so. You must help the compiler choose whether the regular or move version of the function should be called.
In which situations would I want to tell the compiler that it should treat a variable as an rvalue? This will most likely happen in template or library functions, where you know that an intermediate result could be salvaged (rather than allocating a new instance).
std::move itself doesn't really do much. I thought that it called the moved constructor for an object, but it really just performs a type cast (casting an lvalue variable to an rvalue so that the said variable can be passed as an argument to a move constructor or assignment operator).
So std::move is just used as a precursor to using move semantics. Move semantics is essentially an efficient way for dealing with temporary objects.
Consider Object A = B + (C + (D + (E + F)));
This is nice looking code, but E + F produces a temporary object. Then D + temp produces another temporary object and so on. In each normal "+" operator of a class, deep copies occur.
For example
Object Object::operator+ (const Object& rhs) {
Object temp (*this);
// logic for adding
return temp;
}
The creation of the temporary object in this function is useless - these temporary objects will be deleted at the end of the line anyway as they go out of scope.
We can rather use move semantics to "plunder" the temporary objects and do something like
Object& Object::operator+ (Object&& rhs) {
// logic to modify rhs directly
return rhs;
}
This avoids needless deep copies being made. With reference to the example, the only part where deep copying occurs is now E + F. The rest uses move semantics. The move constructor or assignment operator also needs to be implemented to assign the result to A.
"What is it?" and "What does it do?" has been explained above.
I will give a example of "when it should be used".
For example, we have a class with lots of resource like big array in it.
class ResHeavy{ // ResHeavy means heavy resource
public:
ResHeavy(int len=10):_upInt(new int[len]),_len(len){
cout<<"default ctor"<<endl;
}
ResHeavy(const ResHeavy& rhs):_upInt(new int[rhs._len]),_len(rhs._len){
cout<<"copy ctor"<<endl;
}
ResHeavy& operator=(const ResHeavy& rhs){
_upInt.reset(new int[rhs._len]);
_len = rhs._len;
cout<<"operator= ctor"<<endl;
}
ResHeavy(ResHeavy&& rhs){
_upInt = std::move(rhs._upInt);
_len = rhs._len;
rhs._len = 0;
cout<<"move ctor"<<endl;
}
// check array valid
bool is_up_valid(){
return _upInt != nullptr;
}
private:
std::unique_ptr<int[]> _upInt; // heavy array resource
int _len; // length of int array
};
Test code:
void test_std_move2(){
ResHeavy rh; // only one int[]
// operator rh
// after some operator of rh, it becomes no-use
// transform it to other object
ResHeavy rh2 = std::move(rh); // rh becomes invalid
// show rh, rh2 it valid
if(rh.is_up_valid())
cout<<"rh valid"<<endl;
else
cout<<"rh invalid"<<endl;
if(rh2.is_up_valid())
cout<<"rh2 valid"<<endl;
else
cout<<"rh2 invalid"<<endl;
// new ResHeavy object, created by copy ctor
ResHeavy rh3(rh2); // two copy of int[]
if(rh3.is_up_valid())
cout<<"rh3 valid"<<endl;
else
cout<<"rh3 invalid"<<endl;
}
output as below:
default ctor
move ctor
rh invalid
rh2 valid
copy ctor
rh3 valid
We can see that std::move with move constructor makes transform resource easily.
Where else is std::move useful?
std::move can also be useful when sorting an array of elements. Many sorting algorithms (such as selection sort and bubble sort) work by swapping pairs of elements. Previously, we’ve had to resort to copy-semantics to do the swapping. Now we can use move semantics, which is more efficient.
It can also be useful if we want to move the contents managed by one smart pointer to another.
Cited:
https://www.learncpp.com/cpp-tutorial/15-4-stdmove/
std::move itself does nothing rather than a static_cast. According to cppreference.com
It is exactly equivalent to a static_cast to an rvalue reference type.
Thus, it depends on the type of the variable you assign to after the move, if the type has constructors or assign operators that takes a rvalue parameter, it may or may not steal the content of the original variable, so, it may leave the original variable to be in an unspecified state:
Unless otherwise specified, all standard library objects that have been moved from being placed in a valid but unspecified state.
Because there is no special move constructor or move assign operator for built-in literal types such as integers and raw pointers, so, it will be just a simple copy for these types.
Here is a full example, using std::move for a (simple) custom vector
Expected output:
c: [10][11]
copy ctor called
copy of c: [10][11]
move ctor called
moved c: [10][11]
Compile as:
g++ -std=c++2a -O2 -Wall -pedantic foo.cpp
Code:
#include <iostream>
#include <algorithm>
template<class T> class MyVector {
private:
T *data;
size_t maxlen;
size_t currlen;
public:
MyVector<T> () : data (nullptr), maxlen(0), currlen(0) { }
MyVector<T> (int maxlen) : data (new T [maxlen]), maxlen(maxlen), currlen(0) { }
MyVector<T> (const MyVector& o) {
std::cout << "copy ctor called" << std::endl;
data = new T [o.maxlen];
maxlen = o.maxlen;
currlen = o.currlen;
std::copy(o.data, o.data + o.maxlen, data);
}
MyVector<T> (const MyVector<T>&& o) {
std::cout << "move ctor called" << std::endl;
data = o.data;
maxlen = o.maxlen;
currlen = o.currlen;
}
void push_back (const T& i) {
if (currlen >= maxlen) {
maxlen *= 2;
auto newdata = new T [maxlen];
std::copy(data, data + currlen, newdata);
if (data) {
delete[] data;
}
data = newdata;
}
data[currlen++] = i;
}
friend std::ostream& operator<<(std::ostream &os, const MyVector<T>& o) {
auto s = o.data;
auto e = o.data + o.currlen;;
while (s < e) {
os << "[" << *s << "]";
s++;
}
return os;
}
};
int main() {
auto c = new MyVector<int>(1);
c->push_back(10);
c->push_back(11);
std::cout << "c: " << *c << std::endl;
auto d = *c;
std::cout << "copy of c: " << d << std::endl;
auto e = std::move(*c);
delete c;
std::cout << "moved c: " << e << std::endl;
}
std::move simply casts a variable to an rvalue reference. This rvalue reference is notated with &&. Let's say you have a class Foo and you instantiate an object like this
Foo foo = Foo();
If you then write
Foo foo2 = std::move(foo);
that's the same thing as If I wrote
Foo foo2 = (Foo&&) foo;
std::move replaces this cast to an rvalue reference.
The reason why you would want to write either of the previous 2 lines of code
is that if you write
Foo foo2 = foo;
The copy constructor will be called.
Let's say Foo instances have a pointer to some data on the heap which they own.
In Foo's destructor that data on the heap gets deleted.
If you want to distinghuish between copying the data from the heap and taking ownership of that data, you can write a constructor which takes in const Foo& and that constructor can perform the deep copy. Then you can write a constructor which takes in an rvalue reference (Foo&&) and this constructor can simply rewire the pointers.
This constructor which takes in Foo&& will be called when you write
Foo foo2 = std::move(foo);
and when you write
Foo foo2 = (Foo&&) foo;