I know of the following situations in c++ where the copy constructor would be invoked:
when an existing object is assigned an object of it own class
MyClass A,B;
A = new MyClass();
B=A; //copy constructor called
if a functions receives as argument, passed by value, an object of a class
void foo(MyClass a);
foo(a); //copy constructor invoked
when a function returns (by value) an object of the class
MyClass foo ()
{
MyClass temp;
....
return temp; //copy constructor called
}
Please feel free to correct any mistakes I've made; but I am more curious if there are any other situations in which the copy constructor is called.
When an existing object is assigned an object of it own class
B = A;
Not necessarily. This kind of assignment is called copy-assignment, meaning the assignment operator of the class will be called to perform memberwise assignment of all the data members. The actual function is MyClass& operator=(MyClass const&)
The copy-constructor is not invoked here. This is because the assignment operator takes a reference to its object, and therefore no copy-construction is performed.
Copy-assignment is different from copy-initialization because copy-initialization is only done when an object is being initialized. For example:
T y = x;
x = y;
The first expression initializes y by copying x. It invokes the copy-constructor MyClass(MyClass const&).
And as mentioned, x = y is a call to the assignment operator.
(There is also something called copy-elison whereby the compiler will elide calls to the copy-constructor. Your compiler more than likely uses this).
If a functions receives as argument, passed by value, an object of a class
void foo(MyClass a);
foo(a);
This is correct. However, note that in C++11 if a is an xvalue and if MyClass has the appropriate constructor MyClass(MyClass&&), a can be moved into the parameter.
(The copy-constructor and the move-constructor are two of the default compiler-generated member functions of a class. If you do not supply them yourself, the compiler will generously do so for you under specific circumstances).
When a function returns (by value) an object of the class
MyClass foo ()
{
MyClass temp;
....
return temp; // copy constructor called
}
Through return-value optimization, as mentioned in some of the answers, the compiler can remove the call to the copy-constructor. By using the compiler option -fno-elide-constructors, you can disable copy-elison and see that the copy-constructor would indeed be called in these situations.
I might be wrong about this, but this class allows you to see what is called and when:
class a {
public:
a() {
printf("constructor called\n");
};
a(const a& other) {
printf("copy constructor called\n");
};
a& operator=(const a& other) {
printf("copy assignment operator called\n");
return *this;
};
};
So then this code:
a b; //constructor
a c; //constructor
b = c; //copy assignment
c = a(b); //copy constructor, then copy assignment
produces this as the result:
constructor called
constructor called
copy assignment operator called
copy constructor called
copy assignment operator called
Another interesting thing, say you have the following code:
a* b = new a(); //constructor called
a* c; //nothing is called
c = b; //still nothing is called
c = new a(*b); //copy constructor is called
This occurs because when you when you assign a pointer, that does nothing to the actual object.
Situation (1) is incorrect and does not compile the way you've written it. It should be:
MyClass A, B;
A = MyClass(); /* Redefinition of `A`; perfectly legal though superfluous: I've
dropped the `new` to defeat compiler error.*/
B = A; // Assignment operator called (`B` is already constructed)
MyClass C = B; // Copy constructor called.
You are correct in case (2).
But in case (3), the copy constructor may not be called: if the compiler can detect no side effects then it can implement return value optimisation to optimise out the unnecessary deep copy. C++11 formalises this with rvalue references.
This is basically correct (other than your typo in #1).
One additional specific scenario to watch out for is when you have elements in a container, the elements may be copied at various times (for example, in a vector, when the vector grows or some elements are removed). This is actually just an example of #1, but it can be easy to forget about it.
There are 3 situations in which the copy constructor is called:
When we make copy of an object.
When we pass an object as an argument by value to a method.
When we return an object from a method by value.
these are the only situations....i think...
The following are the cases when copy constructor is called.
When instantiating one object and initializing it with values from another object.
When passing an object by value.
When an object is returned from a function by value.
Others have provided good answers, with explanations and references.
In addition, I have written a class to check the different type of instantations/assigments (C++11 ready), within an extensive test:
#include <iostream>
#include <utility>
#include <functional>
template<typename T , bool MESSAGES = true>
class instantation_profiler
{
private:
static std::size_t _alive , _instanced , _destroyed ,
_ctor , _copy_ctor , _move_ctor ,
_copy_assign , _move_assign;
public:
instantation_profiler()
{
_alive++;
_instanced++;
_ctor++;
if( MESSAGES ) std::cout << ">> construction" << std::endl;
}
instantation_profiler( const instantation_profiler& )
{
_alive++;
_instanced++;
_copy_ctor++;
if( MESSAGES ) std::cout << ">> copy construction" << std::endl;
}
instantation_profiler( instantation_profiler&& )
{
_alive++;
_instanced++;
_move_ctor++;
if( MESSAGES ) std::cout << ">> move construction" << std::endl;
}
instantation_profiler& operator=( const instantation_profiler& )
{
_copy_assign++;
if( MESSAGES ) std::cout << ">> copy assigment" << std::endl;
}
instantation_profiler& operator=( instantation_profiler&& )
{
_move_assign++;
if( MESSAGES ) std::cout << ">> move assigment" << std::endl;
}
~instantation_profiler()
{
_alive--;
_destroyed++;
if( MESSAGES ) std::cout << ">> destruction" << std::endl;
}
static std::size_t alive_instances()
{
return _alive;
}
static std::size_t instantations()
{
return _instanced;
}
static std::size_t destructions()
{
return _destroyed;
}
static std::size_t normal_constructions()
{
return _ctor;
}
static std::size_t move_constructions()
{
return _move_ctor;
}
static std::size_t copy_constructions()
{
return _copy_ctor;
}
static std::size_t move_assigments()
{
return _move_assign;
}
static std::size_t copy_assigments()
{
return _copy_assign;
}
static void print_info( std::ostream& out = std::cout )
{
out << "# Normal constructor calls: " << normal_constructions() << std::endl
<< "# Copy constructor calls: " << copy_constructions() << std::endl
<< "# Move constructor calls: " << move_constructions() << std::endl
<< "# Copy assigment calls: " << copy_assigments() << std::endl
<< "# Move assigment calls: " << move_assigments() << std::endl
<< "# Destructor calls: " << destructions() << std::endl
<< "# " << std::endl
<< "# Total instantations: " << instantations() << std::endl
<< "# Total destructions: " << destructions() << std::endl
<< "# Current alive instances: " << alive_instances() << std::endl;
}
};
template<typename T , bool MESSAGES>
std::size_t instantation_profiler<T,MESSAGES>::_alive = 0;
template<typename T , bool MESSAGES>
std::size_t instantation_profiler<T,MESSAGES>::_instanced = 0;
template<typename T , bool MESSAGES>
std::size_t instantation_profiler<T,MESSAGES>::_destroyed = 0;
template<typename T , bool MESSAGES>
std::size_t instantation_profiler<T,MESSAGES>::_ctor = 0;
template<typename T , bool MESSAGES>
std::size_t instantation_profiler<T,MESSAGES>::_copy_ctor = 0;
template<typename T , bool MESSAGES>
std::size_t instantation_profiler<T,MESSAGES>::_move_ctor = 0;
template<typename T , bool MESSAGES>
std::size_t instantation_profiler<T,MESSAGES>::_copy_assign = 0;
template<typename T , bool MESSAGES>
std::size_t instantation_profiler<T,MESSAGES>::_move_assign = 0;
Here is the test:
struct foo : public instantation_profiler<foo>
{
int value;
};
//Me suena bastante que Boost tiene una biblioteca con una parida de este estilo...
struct scoped_call
{
private:
std::function<void()> function;
public:
scoped_call( const std::function<void()>& f ) : function( f ) {}
~scoped_call()
{
function();
}
};
foo f()
{
scoped_call chapuza( [](){ std::cout << "Exiting f()..." << std::endl; } );
std::cout << "I'm in f(), which returns a foo by value!" << std::endl;
return foo();
}
void g1( foo )
{
scoped_call chapuza( [](){ std::cout << "Exiting g1()..." << std::endl; } );
std::cout << "I'm in g1(), which gets a foo by value!" << std::endl;
}
void g2( const foo& )
{
scoped_call chapuza( [](){ std::cout << "Exiting g2()..." << std::endl; } );
std::cout << "I'm in g2(), which gets a foo by const lvalue reference!" << std::endl;
}
void g3( foo&& )
{
scoped_call chapuza( [](){ std::cout << "Exiting g3()..." << std::endl; } );
std::cout << "I'm in g3(), which gets an rvalue foo reference!" << std::endl;
}
template<typename T>
void h( T&& afoo )
{
scoped_call chapuza( [](){ std::cout << "Exiting h()..." << std::endl; } );
std::cout << "I'm in h(), which sends a foo to g() through perfect forwarding!" << std::endl;
g1( std::forward<T>( afoo ) );
}
int main()
{
std::cout << std::endl << "Just before a declaration ( foo a; )" << std::endl; foo a;
std::cout << std::endl << "Just before b declaration ( foo b; )" << std::endl; foo b;
std::cout << std::endl << "Just before c declaration ( foo c; )" << std::endl; foo c;
std::cout << std::endl << "Just before d declaration ( foo d( f() ); )" << std::endl; foo d( f() );
std::cout << std::endl << "Just before a to b assigment ( b = a )" << std::endl; b = a;
std::cout << std::endl << "Just before ctor call to b assigment ( b = foo() )" << std::endl; b = foo();
std::cout << std::endl << "Just before f() call to b assigment ( b = f() )" << std::endl; b = f();
std::cout << std::endl << "Just before g1( foo ) call with lvalue arg ( g1( a ) )" << std::endl; g1( a );
std::cout << std::endl << "Just before g1( foo ) call with rvalue arg ( g1( f() ) )" << std::endl; g1( f() );
std::cout << std::endl << "Just before g1( foo ) call with lvalue ==> rvalue arg ( g1( std::move( a ) ) )" << std::endl; g1( std::move( a ) );
std::cout << std::endl << "Just before g2( const foo& ) call with lvalue arg ( g2( b ) )" << std::endl; g2( b );
std::cout << std::endl << "Just before g2( const foo& ) call with rvalue arg ( g2( f() ) )" << std::endl; g2( f() );
std::cout << std::endl << "Just before g2( const foo& ) call with lvalue ==> rvalue arg ( g2( std::move( b ) ) )" << std::endl; g2( std::move( b ) );
//std::cout << std::endl << "Just before g3( foo&& ) call with lvalue arg ( g3( c ) )" << std::endl; g3( c );
std::cout << std::endl << "Just before g3( foo&& ) call with rvalue arg ( g3( f() ) )" << std::endl; g3( f() );
std::cout << std::endl << "Just before g3( foo&& ) call with lvalue ==> rvalue arg ( g3( std::move( c ) ) )" << std::endl; g3( std::move( c ) );
std::cout << std::endl << "Just before h() call with lvalue arg ( h( d ) )" << std::endl; h( d );
std::cout << std::endl << "Just before h() call with rvalue arg ( h( f() ) )" << std::endl; h( f() );
std::cout << std::endl << "Just before h() call with lvalue ==> rvalue arg ( h( std::move( d ) ) )" << std::endl; h( std::move( d ) );
foo::print_info( std::cout );
}
This is an abstract of the test compiled with GCC 4.8.2 with -O3 and -fno-elide-constructors flags:
Normal constructor calls: 10
Copy constructor calls: 2
Move constructor calls: 11
Copy assigment calls: 1
Move assigment calls: 2
Destructor calls: 19
Total instantations: 23
Total destructions: 19
Current alive instances: 4
Finally the same test with copy elision enabled:
Normal constructor calls: 10
Copy constructor calls: 2
Move constructor calls: 3
Copy assigment calls: 1
Move assigment calls: 2
Destructor calls: 11
Total instantations: 15
Total destructions: 11
Current alive instances: 4
Here is the complete code running at ideone.
Related
I'm working on my own smart pointer and I ran into some weird problems. The move assignment operator was not being called. So I wrote a test class and was able to reproduce the issue. The move assignment operator is not called but a copy assignment occurs (even when there is no copy assignment operator).
This is my test class
#include <utility>
#include <iostream>
struct tag_t {};
constexpr tag_t tag {};
template <typename T>
struct Foo {
Foo() noexcept
: val{} {
std::cout << "Default construct\n";
}
template <typename U>
Foo(tag_t, const U &val) noexcept
: val{val} {
std::cout << "Construct " << val << '\n';
}
~Foo() noexcept {
std::cout << "Destruct " << val << '\n';
}
template <typename U>
Foo(Foo<U> &&other) noexcept
: val{std::exchange(other.val, U{})} {
std::cout << "Move construct " << val << '\n';
}
template <typename U>
Foo &operator=(Foo<U> &&other) noexcept {
std::cout << "Move assign " << other.val << '\n';
val = std::exchange(other.val, U{});
return *this;
}
T val;
};
These are the tests
int main() {
{
Foo<int> num;
std::cout << "Value " << num.val << '\n';
num = {tag, 5};
std::cout << "Value " << num.val << '\n';
}
std::cout << '\n';
{
Foo<int> num;
std::cout << "Value " << num.val << '\n';
num = Foo<int>{tag, 5};
std::cout << "Value " << num.val << '\n';
}
return 0;
}
After running the tests, I get these results
Default construct
Value 0
Construct 5
Destruct 5
Value 5
Destruct 5
Default construct
Value 0
Construct 5
Move assign 5
Destruct 0
Value 5
Destruct 5
What baffles me is the output of the first test. The move assignment operator is not called but a copy assignment takes place. This results in 5 being destroyed twice. Not ideal when you're trying to make a smart pointer!
I'm compiling with Apple Clang with optimizations disabled. Can someone explain my observations? Also, how do I ensure that the move assignment operator is called in the first test?
template <typename U>
Foo &operator=(Foo<U> &&other) noexcept;
this cannot be called by ={ }.
Instead, Foo& operator=(Foo&&)noexcept is called.
Template methods are never special member functions. Explicitly default, delete or implement them.
This question already has answers here:
const reference to a temporary object becomes broken after function scope (life time)
(2 answers)
Closed 4 years ago.
For regular local const reference variables, the scope is prolonged. Which is why the following code works as expected:
#include <iostream>
#include <memory>
struct foo
{
foo()
{
std::cout << "foo() #" << (void*)this << std::endl;
}
~foo()
{
std::cout << "~foo() #" << (void*)this << std::endl;
}
};
int main()
{
auto const& f = std::make_shared<foo>();
std::cout << "f = " << f.get() << std::endl;
return 0;
}
// prints:
// foo() #0x55f249c58e80
// f = 0x55f249c58e80
// ~foo() #0x55f249c58e80
It seems though that this does not work as expected when assigning a moved object using std::move():
#include <iostream>
#include <memory>
#include <list>
struct foo
{
foo()
{
std::cout << "foo() #" << (void*)this << std::endl;
}
~foo()
{
std::cout << "~foo() #" << (void*)this << std::endl;
}
};
int main()
{
std::list<std::shared_ptr<foo>> l;
l.push_back(std::make_shared<foo>());
auto const& f = std::move(l.front());
l.clear();
std::cout << "f = " << f.get() << std::endl;
return 0;
}
// prints
// foo() #0x564edb58fe80
// ~foo() #0x564edb58fe80
// f = 0x564edb58fe80
Does std::move() indeed change the scope, or am I dealing with a compiler bug?
Changing the variable from auto const& f to just auto f fixes the problem. If I wrap the move into another function, it also works:
auto const& f = [&]() { return std::move(l.front()); }();
It's almost like std::move() does not share the same semantics as a function call, but rather as if it was just a regular variable assignment:
auto const& f = std::move(l.front());
Let's put aside std::move() and create this function:
struct sometype {};
const sometype &foobar( const sometype &cref ) { return cref; }
and now we use it:
const sometype &ref = foobar( sometype() );
What do you think, will lifetime of temporary be prolongated in this case? No it would not. Lifetime is only prolongated when you assign to a reference directly. When it goes through a function or through static_cast or std::move that prolongation is gone. So you have exactly the same issue with std::move()
struct sometype {
sometype() { std::cout << "ctype()" << std::endl; }
~sometype() { std::cout << "~ctype()" << std::endl; }
};
const sometype &foobar( const sometype &cref ) { return cref; }
int main()
{
const sometype &cref1 = foobar( sometype() );
sometype &&cref2 = std::move( sometype() );
std::cout << "main continues" << std::endl;
}
output:
ctype()
~ctype()
ctype()
~ctype()
main continues
live example
Note: you should not use std::move() on return statement, it does not give you anything.
For your code change. You should remember that std::move() does not move anything. It is done by special assignment operator or constructor (if they provided for a type). So when you write this code:
const type &ref = std::move( something );
there is no constructor nor assignment operator involved and so no moving happens. For actual moving to happen you have to assign it to a variable:
type val = std::move( something );
now moving would happen if possible or copy otherwise.
The following code snippet which I was writing to understand move CTOR behaviour is giving me hard time to understand it's output:
#include <iostream>
class Temp
{
public:
Temp(){
std::cout << "Temp DEFAULT CTOR called" << std::endl;
mp_Val = nullptr;
}
Temp(int inp) {
std::cout << "Temp CTOR called" << std::endl;
mp_Val = new int(inp);
}
Temp(const Temp& inp) {
std::cout << "Temp COPY CTOR called" << std::endl;
mp_Val = new int(*inp.mp_Val);
}
Temp& operator= (const Temp& inp) {
std::cout << "Temp ASSIGNMENT OPER called" << std::endl;
mp_Val = new int(*inp.mp_Val);
return *this;
}
int* mp_Val;
};
class B
{
public:
B(){
std::cout << "Class B DEFAULT CTOR" << std::endl;
mp_Val = nullptr;
}
B(int inp) {
std::cout << "Class B CTOR" << std::endl;
mp_Val = new Temp(inp);
}
B(const B& in) {
std::cout << "Class B COPY CTOR" << std::endl;
mp_Val = in.mp_Val;
}
B(B&& in){
std::cout << "Class B MOVE CTOR" << std::endl; //Doubt: 1
}
Temp *mp_Val;
};
int main() {
B obj1(200);
B obj2 = std::move(obj1);
auto temp = obj1.mp_Val;
std::cout << "Obj1 B::Temp address: " << obj1.mp_Val << std::endl;
std::cout << "Obj2 B::Temp address: " << obj2.mp_Val << std::endl; //Doubt: 2
return 0;
}
Output:
Class B CTOR
Temp CTOR called
Class B MOVE CTOR
Obj1 B::Temp address: 0xd48030
Obj2 B::Temp address: 0x400880
GCC version: 4.6.3
My question is about the line marked as Doubt 2. Should not the address be printed as 0? As per my understanding, as I have defined an empty move CTOR (marked as Doubt 1) in class B, it should call the default CTOR of class Temp (which it's not calling as evident from the logs) to initialise its member variable mp_Val of type Temp.
There is obviously something that I am missing.
As per my understanding, as I have defined an empty move CTOR (marked as Doubt 1) in class B, it should call the default CTOR of class Temp (which it's not calling as evident from the logs) to initialise its member variable mp_Val of type Temp.
Your member variable isn't of type Temp, it's of type Temp *. You're right that the lack of an initialiser means that that member will be default-constructed, and for type Temp that would involve calling the default constructor. However, for pointer types, default construction leaves the object uninitialised.
I am studying constructor, copy constructors and copy assignment a while and I read a fair amount of contents mainly from here as In which situations is the C++ copy constructor called? and What is The Rule of Three? and others, but I still no deeply understanding their exactly behavior. I am performing some tests with the class Boo and Foo bellow and their results are not the ones that I was expecting. In the class Boo I explictly declared its default constructor, copy constructor and copy assingment, and I defined them to do nothing. From this I was expecting that the objects from this class would have member variables with random values, what was indeed observed. Nevertheless, this behavior was not seen with the objects from the class Foo. The only difference between these classes are that the later is a singleton, but its default constructor, copy constructor and copy assignment still explicit declared and defined to do nothing. From this I state my question: Why the values of the member variables of objects from class Foo are not randomly initialized, but always has de same value of "0" and " "?
#include <iostream>
class Boo {
private:
int x;
char y;
public:
int getInt() { return x; }
char getChar() { return y; }
Boo () { std::cout << "Boo default constructor\n"; }
Boo ( const Boo& other ) { std::cout << "Boo copy constructor\n"; }
Boo& operator= ( const Boo& other) { std::cout << "Boo copy assinment\n";
return *this;}
};
class Foo {
private:
int x;
char y;
static Foo *instance;
protected:
Foo() { std::cout << "Foo default constructor\n"; }
Foo ( const Foo& other ) { std::cout << "Foo copy constructor\n"; }
Foo& operator=( const Foo& other) { std::cout << "Foo copy assignment\n"; }
public:
static Foo & uniqueInst();
int getInt() { return x; }
char getChar() { return y; }
};
Foo *Foo::instance = 0;
Foo & Foo::uniqueInst(){
if(!instance) instance = new Foo();
return *instance;
};
int main(){
Boo b1; // default constructor
Boo b2; // default constructor
Boo b3 = b1; // copy constructor
b2 = b1;
std::cout << b1.getInt() << std::endl; // Random values since the constructor does nothing
std::cout << b1.getChar() << std::endl;
std::cout << b2.getInt() << std::endl; // Random values since the copy assignment does nothing
std::cout << b2.getChar() << std::endl;
std::cout << b3.getInt() << std::endl; // Random values since the copy constructor does nothing
std::cout << b3.getChar() << std::endl;
Foo *foo;
foo = &Foo::uniqueInst(); // defaulf construtor
std::cout << foo << std::endl;
std::cout << foo->getInt() << std::endl; // Why not random values?
std::cout << foo->getChar() << std::endl;
};
Say I have this function:
template <class A>
inline A f()
{
A const r(/* a very complex and expensive construction */);
return r;
}
Is it a good idea to declare r const, since a const variable cannot be moved? Note that the returned value is not const. The qualm I am grappling is, that r truly is const, but it may not be a good idea to declare it as such. Yet the qualifier should be helping the compiler generate better code.
As demonstrated here, NRVO elides the copy of r implied by the line return r;
#include <iostream>
struct A {
const char* name;
A( const char* name_ ):name(name_) { std::cout << "created " << name << "\n"; }
A(A const&){ std::cout << "copied " << name << "\n"; }
A(A &&){ std::cout << "moved " << name << "\n"; }
};
A f() {
std::cout << "start of f()\n";
A const r("bob");
std::cout << "body of f()\n";
return r;
}
int main() {
A x = f();
}
And the copy in main is also elided.
If you block NRVO and RVO in some other way (for instance using the flag -fno-elide-constructors when compiling with GCC), the const can cause your object to be copied instead of moved. You can see this if we remove the copy constructor from A:
#include <iostream>
struct A {
const char* name;
A( const char* name_ ):name(name_) { std::cout << "created " << name << "\n"; }
//A(A const&){ std::cout << "copied " << name << "\n"; }
A(A &&){ std::cout << "moved " << name << "\n"; }
};
A f() {
std::cout << "start of f()\n";
A const r("bob");
std::cout << "body of f()\n";
return r;
}
int main() {
A x = f();
}
the code no longer compiles. While the copy constructor isn't executed so long as NRVO occurs, its existence is required by your const local variable.
Now, NRVO requires a few things, such as a single variable which is returned along every single execution path of the function in question: if you ever "abort" and do a return A(), NRVO is blocked, and your const local variable suddenly forces a copy at all return sites.
If class A is under your control, and you want to return const objects by move, you can do
mutable bool resources_were_stolen = false;
and set that to true in a const move constructor
A(const A&& other) { ...; other.resources_were_stolen = true; }
~A() { if (!resources_were_stolen) ... }
Actually, the destructor probably would become if (resources_were_stolen) some_unique_ptr.release();, using the fact that objects lose their const-ness during construction and destruction.