In the code show below, how do I assign rvalue to an object A in function main?
#include <iostream>
using namespace std;
class A
{
public:
int* x;
A(int arg) : x(new int(arg)) { cout << "ctor" << endl;}
A(const A& RVal) {
x = new int(*RVal.x);
cout << "copy ctor" << endl;
}
A(A&& RVal) {
this->x = new int(*RVal.x);
cout << "move ctor" << endl;
}
~A()
{
delete x;
}
};
int main()
{
A a(8);
A b = a;
A&& c = A(4); // it does not call move ctor? why?
cin.ignore();
return 0;
}
Thanks.
Any named instance is l-value.
Examples of code with move constructor:
void foo(A&& value)
{
A b(std::move(value)); //move ctr
}
int main()
{
A c(5); // ctor
A cc(std::move(c)); // move ctor
foo(A(4));
}
Related
I meet a course programming problem, which asks me to initialize the A a using passing by reference (initialize the A a in the func). How can I call A's constructor by A's reference?
#include <iostream>
using namespace std;
class A
{
public:
int x;
A()
{
cout << "default constructor" << endl;
x = 1;
}
A(int x)
{
cout << "constructor with param = " << x << endl;
this->x = x;
}
~A() {
cout << "destructor" << endl;
}
void print() {
cout << x << endl;
}
};
void fun(A& a)
{
a.A::A(10); // error!
return;
}
int main()
{
A a;
fun(a);
a.print();
return EXIT_SUCCESS;
}
There is a background of this problem. The teacher want us to replicate the NRVO(named return value optimization) result.
#include <iostream>
using namespace std;
class A
{
public:
int x;
A()
{
cout << "default constructor" << endl;
x = 1;
}
A(int x)
{
cout << "constructor with param = " << x << endl;
this->x = x;
}
~A() {
cout << "destructor" << endl;
}
void print() {
cout << x << endl;
}
};
A fun() {
A a = A(10);
return a;
}
int main()
{
A a = fun();
return EXIT_SUCCESS;
}
default g++ compiler:
constructor with param = 10
destructor
if we close the NRVO:
g++ test.cpp -fno-elide-constructors
constructor with param = 10
destructor
destructor
destructor
destructor
The teacher want us to replicate the NRVO(named return value optimization) result by passing by reference.
The syntax a.A::A(10); is incorrect.
Constructor is used to create an object of a class, you cannot call it on an already existing object. Even a constructor cannot be explicitly called. It is implicitly called by the compiler.
From general-1.sentence-2:
Constructors do not have names.
Thus, you cannot call a constructor explicitly. The compiler will automatically call the constructor when an object of that class-type is created.
You can not, not like this.
A reference always points to an initialized object. So you already failed before you called the function. The "return" argument is already initialized. And you can't initialized an initialized value again, not legally.
You can cheat by calling
std::construct_at(&a, 10);
For it to really reflect NRVO you could have something like this:
void fun(A *a)
{
std::construct_at(a, 10);
}
union UninitializedA {
std::byte uninitialized[sizeof(A)];
A a;
};
int main()
{
UninitializedA u;
fun(&u.a);
u.a.print();
u.a.~A();
return EXIT_SUCCESS;
}
#include<iostream>
using namespace std;
class test
{
public:
int a,b;
test()
{
cout<<"default construictor";
}
test(int x,int y):a(x),b(y){
cout<<"parmetrized constructor";
}
};
int main()
{
test t;
cout<<t.a;
//t=(2,3);->gives error
t={2,3}; //calls paramterized constructor
cout<<t.a;
}
Output:- default construictor4196576parmetrized constructor2
why in case of the above example, parameterized constructor(even though default constructor is already called.) is called in case of {} and not in ()
I added some additional code to show what is actually happening.
#include<iostream>
using namespace std;
class test
{
public:
int a,b;
test()
{
cout << "default constructor" << endl;
}
~test()
{
cout << "destructor" << endl;
}
test(int x,int y):a(x),b(y)
{
cout << "parameterized constructor" << endl;
}
test& operator=(const test& rhs)
{
a = rhs.a;
b = rhs.b;
cout << "assignment operator" << endl;
return *this;
}
};
int main()
{
test t;
cout << t.a << endl;
//t=(2,3);->gives error
t={2,3}; //calls parameterized constructor
cout << t.a << endl;
}
Output:
default constructor
4197760
parameterized constructor
assignment operator
destructor
2
destructor
So the statement t={2,3}; is actually constructing a new test object using the parameterized constructor, calling the assignment operator to set t to be equal to the new, temporary test object, and then destroying the temporary test object. It's equivalent to the statement t=test(2,3).
use
test t(2,3);
instead of
test t;
t=(2,3);
Because parentheses are used after object declaration.
This code is calling another ctor in one ctor:
#include <iostream>
using namespace std;
class F {
public:
F() { cout << "ctor1\n"; }
F(int) { cout << "ctor2\n"; }
~F() { cout << "dtor\n"; }
};
class Foo {
F f;
public:
Foo() : f() { cout << "1\n"; }
Foo(int i) : f(i) { Foo(); cout << "2\n"; }
};
int main() {
Foo object(1);
return 0;
}
The result is:
ctor2
ctor1
1
dtor
2
dtor
It seems the member variable f destroyed twice here, is it Okay?
Here
Foo(int i) { Foo(); cout << "2\n"; }
You are not using delegating constructor. What you're doing is creating a temporary instance of Foo in the constructor body (and destroying it immediately).
The correct syntax for delegating constructor is
Foo(int i) : Foo() { cout << "2\n"; }
I was writing this program where I defined a class X and manually defined its constructors and destructor so that I could have a print statement in each of them and see when they are called.
However, the problem seems to be with my definition of the copy constructor.
It gives the following error:
warning: passing const X as this argument of int X::getI() discards const
What's the cause of this error?
Code snippet of class:
class X {
public:
X() {
cout << "Default Constructor called\n";
i = 0;
}
X(int i) {
cout << "Parameterized Constructor called\n";
this->i = i;
}
X(const X& x) {
cout << "Copy Constructor called\n";
i = x.getI();
}
~X() {
cout << "Destructor called\n";
}
int getI() {
return i;
}
private:
int i;
};
You are attempting to call the non-const member function getI through a const reference. This is not allowed. Since getI doesn't modify this object, it should be declared as const.
int
getI() const
{
return this->i;
}
Then you'll be able to call it even through a const reference.
getI() not a const member function. You are not allowed to call it on objects that are const. In the copy constructor, x is a const object. Hence, you can't call
x.getI();
Change getI() to a const member function.
int getI() const {
return i;
}
#include <iostream>
using namespace::std;
class X {
public:
X() {
cout << "Default Constructor called\n";
i = 0;
}
X(int i) {
cout << "Parameterized Constructor called\n";
this->i = i;
}
X(const X& x) {
cout << "Copy Constructor called\n";
i = x.getI();
}
~X() {
cout << "Destructor called\n";
}
int getI() const {
return i;
}
private:
int i;
};
int main()
{
}
Here is the correct code, u have to make getI as const
#include <iostream>
using namespace std;
struct A
{
A() {}
A(const A &a) {
cout << "copy constructor" << endl;
}
A& operator=(const A &a) {
cout << "assigment operator" << endl;
}
A(A &&a) {
cout << "move" << endl;
}
A& operator=(A &&a) {
cout << "move" << endl;
}
};
struct B {
A a;
};
B func() {
B b;
return b;
}
int main() {
B b = func();
}
This prints "copy constructor".
For class B the move constructor and the move assignment operator should be automatic generated correct? But why is it using the copy constructor of class A and not the move constructor?
For me it doesn't print anything at all because the copy/move has been elided. However if I thwart RVO with something like:
extern bool choice;
B func() {
B b1, b2;
if (choice)
return b1;
return b2;
}
Then it prints:
move
It may be that your compiler does not yet implement the automatic generation of the move members.