Related
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;
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;
#include <iostream>
#include <vector>
class myClass {
public:
double a;
double& ref;
myClass() : ref(a)
{
a = 1;
}
~myClass() {}
};
using namespace std;
int main()
{
vector<myClass> myVector;
int nIter = 5;
while (nIter--) {
myVector.push_back(myClass());
}
return 0;
}
Hi.
I have myClass and I would like to push_back the myClasses and bring them together in one vector.
But unfortunately, I have to use a reference in myClass.
The problem is when the temporary object destructs, the reference becomes invalid and vector contains the object whose reference is invalidated.
After investigating, I was able to see that those reference variables are pointing at (referencing) the same memory.
I would like to find the way where each vector's element's reference member variable references each vector's element's a(member variable).
Is there any way to achieve this..?
Addition
I would like to describe my situation further.
I have one middle-sized project. In there, the users have the option to choose which variable among member variables will be used in my algorithm. so I made the script in which ref variable is used so that it can change according to the option.
I hope my explanation was clear.
You're getting this behavior because you are initializing your 'ref' member to a value that is on the stack, then the default copy constructor is copying that into the vector.
For example, in my debugger the value I have for ref is:
+ &myVector[1].ref 0x00eff80c {2.0000000000000000} double *
+ &myVector[1] 0x00126940 {a=2.0000000000000000 ref=2.0000000000000000 }
+ myVector { size=0x00000002 } std::vector<myClass,std::allocator<myClass> >
+ &nIter 0x00eff8f0 {0xffffffff} int *
You can see that myVector[1].ref is not inside myVector[1] as you'd expect, and is, in fact, on the stack. You can see that nIter and ref are only 57 bytes apart:
&nIter - (int*)&myVector[0].ref 57
If you want to see how this is implicitly happening you can delete your copy constructor:
myClass(myClass const &rhs) = delete;
inside myClass and you'll get an error at push_back.
Another option is to write your own copy constructor:
myClass(myClass const &rhs) : ref(a) {
a = rhs.a;
}
If you debug this you'll see that the values are correct, and that the memory locations of each ref are now inside the bounds of the myClass objects.
Finally you might be able to use emplace_back instead of push_back, which will construct myClass directly in the vector's memory instead of calling the copy ctor, though I wouldn't recommend this as it leaves this ref copy bug.
also don't forget the assignment operator if you go the copy ctor route:
https://en.wikipedia.org/wiki/Rule_of_three_(C%2B%2B_programming)
e.g.
myClass baz;
baz = myVector[0];
this will invoke operator= and not the copy ctor. My compiler (visual studio c++ latest) automatically deletes operator= if you declare a copy ctor, so it would catch this, but your compiler may not.
I have this problem:
void foo(vector<int> &a){
vector<int> b;
b.push_back(1); // in general many push backs
a = b;
}
since b is a local variable it will be deleted when foo ends.
Will a retain the values inserted in b?
I tried and it does but maybe is only due to chance.
Thank you for your help.
EDIT:
I think this answers my question. Isn't it?
http://www.cplusplus.com/reference/vector/vector/operator=/
So basically a will retain the values in c++98. But I cannot fully understand what it does in c++11.
a = b will invoke the copy assignment operator of std::vector. This operator copies all the contents of the vector from b to a so b can be safely deleted afterwards.
Considering your edit:
This answer is also true for C++11. C++11 only adds a move assignment operator, but this operator is not invoked since b is not an r-value.
Even if b was an r-value (e.g., by wrapping b into a std::move(b)), then the move assignment operator would be invoked, which would also be fine. It would move the contents of b to a, so b would be an empty vector afterwards which can be safely deleted, as well.
What happens in C++ 11 is that the assignment operator will do a copy unless you explicitly tell it to move using std::move.
To force the move you could do this:
void foo(vector<int> &a){
vector<int> b;
b.push_back(1); // in general many push backs
a = std::move(b);
}
For this to work, vector has implemented the assignment operator for move something like this:
vector<T>& vector<T>::operator =(vector<T>&& to_move) {}
When using std::move, this tells the compiler to use the overloaded function that has implemented a move, otherwise it will fall back to const vector<T>& on older versions that have not implemented move semantics.
Move might or might not be faster depending on how this is implemented, but for vector you would expect the internal pointer to be moved from the old vector to the new vector, avoiding a copy of all contained objects.
e.g.
vector<T>& vector<T>::operator =(vector<T>&& to_move)
{
delete [] m_ptrs;
m_ptrs = to_move.m_ptrs;
m_size = to_move.m_size;
to_move.m_ptrs = nullptr;
to_move.m_size = 0;
return *this;
}
This is of course an educated guess, but hopefully it demonstrates what might happen when using move semantics.
If those values are not pointers, they will be copied in a = b so they will be preserved (They will be also previously copied in b when they are inserted http://www.cplusplus.com/reference/vector/vector/push_back/).
If they are, they will be also copied too but if they were pointing to local variables you will get a good collection of invalid pointers.
If, on the other hand, they were pointers to dynamically allocated memory you would not find any problem.
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;