Using address of variable inside constructor of class being reassigned [duplicate] - c++

This question already has answers here:
What is The Rule of Three?
(8 answers)
Closed 4 years ago.
I have been debugging an issue involving an std::function (created from std::bind) where the issue boils down to a more fundamental question.
An example follows below, but in short it's about the case when a class in one way or another stores a pointer to itself (or an instance variable) in its constructor and the class itself is then reassigned. From my experiments it seems the stored pointer will then no longer point to the class (or its instance variable). I wonder what are the best practices to avoid these issues?
An example:
#include <iostream>
class Interface
{
public:
Interface() {};
virtual ~Interface() {};
virtual void callback() = 0;
};
class B
{
public:
B() : interface(NULL) {};
virtual ~B() {};
void set_interface(Interface* interface)
{
this->interface = interface;
}
void print()
{
std::cout << "B is at: " << this << std::endl;
std::cout << " Interface (A) from B is at: " << interface << std::endl;
}
private:
Interface* interface;
};
class A : public Interface
{
public:
A()
{
b.set_interface(this);
};
virtual ~A() {};
virtual void callback()
{
std::cout << "Callback" << std::endl;
}
void print()
{
std::cout << "A is at: " << this << std::endl;
b.print();
}
private:
B b;
};
int main(int argc, char** argv)
{
A a;
a.print();
std::cout << "------" << std::endl;
a = A();
a.print();
return 0;
}
When I run this on my machine I get:
A is at: 0x7ffdb8910af0
B is at: 0x7ffdb8910af8
Interface (A) from B is at: 0x7ffdb8910af0
------
A is at: 0x7ffdb8910af0
B is at: 0x7ffdb8910af8
Interface (A) from B is at: 0x7ffdb8910b10
So, at first everything is fine. The pointer to Interface from B points to the address of A. However, after reassigning (a=A()) the pointer to Interface no longer points to A! I realize this is because as the constructor of A is running to create the second instance, it is in a new memory location and stores the interface pointer to this new memory location. When a is then reassigned the object is moved in to the old location which invalidates the pointer to Interface.
This seems like a common use case to me. Should one always avoid referencing memory in constructors of objects?
In my case I would like to bind a callback function and pass it to a class member, it seems I cannot do it in the constructor then and must do it at a later time?

A is at fault:
A()
{
b.set_interface(this);
};
You define a constructor for A which handles a resource: itself. According to the rule of 0/3/5, A should have a user-defined copy constructor and an assignment operator, and if it makes sense, their move conterparts.
A()
{
b.set_interface(this);
};
A(A const&)
{
b.set_interface(this);
}
A& operator=(A const&)
{
b.set_interface(this);
return *this;
}

Related

Calling a class with an abstract base class from a derived class --safely

I'd like to call the class Foo which has the abstract class Base in its ctor. I'd like to be able to call Foo from Derived which is derived from Base and use Derived's overriding methods rather than Base's.
I'm only able to do this by using a raw pointer as indicated. Is there any way to do this without raw pointers? I tried std::shared_ptr but the compiler complains about abstract classes. Or perhaps is there a better way?
#include <iostream>
class Base {
public:
Base() {
std::cout << "Hello from Base." << std::endl;
}
virtual void show() const = 0;
};
class Foo {
public:
explicit Foo(const Base *s) { // can I avoid this raw pointer?
std::cout << "Hello from Foo." << std::endl;
s->show();
}
Base *s;
};
class Derived : public Base {
public:
Derived() : Base() {
std::cout << "Hello from Derived." << std::endl;
Foo(this);
}
void show() const override {
std::cout << "Hi, I'm Derived::show()." << std::endl;
}
};
int main() {
Derived();
return EXIT_SUCCESS;
}
which produces the following output:
Hello from Base.
Hello from Derived.
Hello from Foo.
Hi, I'm Derived::show().
The code can be rewritten with const reference to Base as
#include <iostream>
class Base {
public:
Base() {
std::cout << "Hello from Base." << std::endl;
}
virtual void show() const = 0;
};
class Foo {
public:
explicit Foo(const Base& b) : s(b) { // member initialization list to set s
std::cout << "Hello from Foo." << std::endl;
s.show();
}
const Base& s;
};
class Derived : public Base {
public:
Derived() : Base() {
std::cout << "Hello from Derived." << std::endl;
Foo(*this); // the parameter would be the object itself *this, instead of a pointer this
}
void show() const override {
std::cout << "Hi, I'm Derived::show()." << std::endl;
}
};
int main() {
Derived();
return EXIT_SUCCESS;
}
A reference should be initialized in a member initializer list of the constructor.
When using a raw pointer or a reference 'Foo' should not own 'Base', i.e. Foo is not responsible for destroying Base and the lifetime of Base should be guaranteed by the owner of Foo.
You have to make sure that Base (=Derived in this case) outlives Foo. That is guaranteed, if the Foo object is owned by Derived, e.g. as member or local variable. Then before Base=Derived is destroyed, Foo is destroyed.
You can use normal references instead of const references, but then the same for the constructor parameter as well as the member variable.
A raw pointer (in comparison to a reference) is idiomatic in cases,
where the parameter can also be the nullptr instead of a valid object or
where the pointer is exchanged with another one during the lifetime of the object.
The first case could be handled by std::optional instead, the second one with assignment of a lightweight object ('view'), which basically encapsulates a pointer or a reference.
So very few cases (e.g. low-level code, data-structures or for compatibility with C) are left, where raw pointers would still be used in modern C++. And even in those cases, having a wrapper object, which just stores a reference as member variable, would have the same performance (and in practice also the same memory layout) as raw pointers, but are much more clean and safe to use.
In some cases, you would prefer a raw pointer to std::optional for performance reasons, when execution speed or memory size really matters. As alternative, a reference to nullptr is not allowed in C++.

Calling virtual method from virtual destructor in C++ [duplicate]

This question already has answers here:
Calling virtual method from destructor - workaround?
(4 answers)
Closed 5 years ago.
I want to destruct an object of class B.
class A {
public:
A() {
std::cout << "construct A" << av::endl;
a = new int;
}
virtual ~A() {
std::cout << "destruct A" << av::endl;
this->clear();
}
virtual void clear() {
std::cout << "clear A" << av::endl;
delete a;
}
protected:
int *a;
};
class B : public A {
public:
B() {
std::cout << "construct B" << av::endl;
b = new int;
}
~B() {
std::cout << "destruct B" << av::endl;
}
void clear() override {
std::cout << "clear B" << av::endl;
delete b;
delete this->a;
}
private:
int *b;
};
And I want it to be done with clear() method. But when I execute following code:
A *a = new B();
delete a;
I get:
construct A construct B destruct B destruct A clear A
And clear B is never printed.
What am I doing wrong?
Informally, In ~A(); the B part is already destroyed, calling any function of B doesn't make any sense.
Effective C++ Item 9: Never call virtual functions during construction or destruction.
Once a derived class destructor has run, the object’s derived class
data members assume undefined values, so C++ treats them as if they no
longer exist. Upon entry to the base class destructor, the object
becomes a base class object, and all parts of C++ — virtual functions,
dynamic_cast s, etc., —treat it that way.

Object not deleted before new is assigned

I'm kind of confused, because I was sure this should work different. Take a look at this code example:
#include <iostream>
#include <string>
using namespace std;
class base
{
public:
virtual ~base() = default;
};
class derived : public base
{
private:
int a = 0;
int *b = nullptr;
std::string lol;
public:
derived(std::string s) : b(new int(6)), lol{s} { cout << "ctor " << lol << endl; }
derived(derived const& d) : lol{d.lol + " copy"} {cout << "copy " << lol << endl; }
virtual ~derived() { cout << "dtor " << lol << endl; delete b; }
virtual void superFunction() { cout << "OMG " << lol << endl; }
};
int main()
{
derived a("a");
derived b("b");
a = b;
}
And the program output with all optimizations off is:
ctor a
ctor b
dtor b
dtor b
I was sure that in this case compiler should generate code that deletes object a and uses copy constructor to create new object. Instead it uses operator= that it implicitly declares.
Can someone explain why? Or point me to C++ standard.
Thanks.
When you write a = b;, compiler calls assignment operator, which will be automatically generated if not present in the code and not marked as deleted. Copy constructor is used only if you try to initialize a new object from another object like this:
derived a("a");
derived b = a;
Also, your code crashes before main returns as it tries to delete b, which points to the same memory from a and from b after a = b; default-assignment.
If you want to delete a with derived destructor after a = b; execution, all you need is copy-and-swap idiom. What is the copy and swap idiom? has a great answer on how to do that in legacy and modern C++. Proper implementation of rule-of-four from that answer will perfectly fit the DRY principle and help you to avoid memory issues. Note the fabulous trick with passing parameter to operator= by value, which makes compiler select the appropriate constructor (copy or move) and allows you to write only four methods instead of all five of them.

C++: Virtual methods [duplicate]

This question already has answers here:
What is object slicing?
(18 answers)
Closed 9 years ago.
I have the following piece of code (#includes and using namespace std omitted):
class A {
public:
void call(){callme();}
private:
virtual void callme() {cout << "I'm A" << endl;}
};
class B : public A {
private:
virtual void callme() {cout << "I'm B" << endl;}
};
class C : public B {
public:
virtual void callme(){ cout << "I'm C" << endl;}
};
int main(){
vector<A> stuff = {
A(), B(), C(),
};
stuff[0].call(); // output: I'm A
stuff[1].call(); // output: I'm A
stuff[2].call(); // output: I'm A
return 0;
}
As stated in the comments, the output of the above program is:
I'm A
I'm A
I'm A
However, I would like that C++ automatically recognizes the type with which the corresponding element was created. I.e. I would like that C++ outputs
I'm A
I'm B
I'm C
(That is, the compiler shall pick the proper subclass for me.)
Is this possible in this scenario (i.e. if all the elements come out of a vector)?
Member functions virtuality works only when you call them from pointer to the actual object, not from an object itself, because in your example objects were automatically statically upcasted to class A. Change your code to:
std::vector<std::unique_ptr<A>> stuff = {
std::unique_ptr<A>(new A()),
std::unique_ptr<A>(new B()),
std::unique_ptr<A>(new C()),
};
stuff[0]->call();
stuff[1]->call();
stuff[2]->call();
For C++ Polymorphism, you should use either pointer or reference. You could do like this
int main(){
vector<A*> stuff;
stuff.push_back(new A);
stuff.push_back(new B);
stuff.push_back(new C);
stuff[0]->call(); // output: I'm A
stuff[1]->call(); // output: I'm A
stuff[2]->call(); // output: I'm A
while (!stuff.empty()){
delete stuff.back();
stuff.pop_back();
}
return 0;
}
Reference: http://www.cplusplus.com/doc/tutorial/polymorphism/

Base class destructor is not virtual and child class destructor is virtual, program crash

Why does the following program crash? I have a base class whose destructor is not virtual but the child class destructor is virtual:
#include <iostream>
class Base {
public:
Base() {
std::cout << "Base::Base CTOR " << std::endl;
}
~Base() {
std::cout << "Base::Base DTOR " << std::endl;
}
};
class Child : public Base {
public:
Child(){
std::cout << "Child::Child CTOR " << std::endl;
}
virtual ~Child() {
std::cout << "Child::Child DTOR " << std::endl;
}
};
int main (int argc, char **argv) {
Base *ptr = new Child;
delete ptr;
}
What you are observing is called "undefined behavior". Make Base's dtor virtual if you want do call delete on Child instance through Base pointer.
From the 2003 standard, 5.3.5/3:
In the first alternative (delete object), if the static type of the
operand is different from its dynamic type, the static type shall be a
base class of the operand’s dynamic type and the static type shall
have a virtual destructor or the behavior is undefined.
You have undefined behavior because the static type of the pointer operand to delete does not match the dynamic type of the object that it points to and you don't meet the requirements for the exception to this rule that allows passing a pointer to a base class to the object being deleted because this exception requires the base class to have a virtual destructor.
Any behaviour is possible including the code working "as expected" or a crash.
Hope this example helps you get the point:
#include <iostream>
class Base {
public:
Base() {
std::cout << "Base::Base CTOR " << std::endl;
}
~Base() {
std::cout << "Base::Base DTOR " << std::endl;
}
private:
protected:
};
class Child : public Base {
public:
Child(){
std::cout << "Child::Child CTOR " << std::endl;
}
~Child(){
std::cout << "Child::Child DTOR " << std::endl;
}
private:
protected:
};
class gChild : public Child {
public:
gChild(){
std::cout << "Child::Child gCTOR " << std::endl;
}
~gChild(){
std::cout << "Child::Child gDTOR " << std::endl;
}
private:
protected:
};
int main ( int argc, char **argv) {
Base *ptr = new gChild;
delete ptr;
}
if virtual ~Base() ,then all destructors' print gets printed.
if virtual ~child() or virtual ~gChild(),only base destructor gets printed.
It's because the destructors executes in opposite direction.and here behaviour is undefined.You must define the base destructor virtual to get the expected result.
Thanks.
Just look at this:
#include <iostream>
class Base
{
public:
void nonvirtualmethod()
{ std::cout << "Base nonvirtualmethod" << std::endl; }
virtual void virtualmethod()
{ std::cout << "Base virtualmethod" << std::endl; }
};
class Derived: public Base
{
public:
void nonvirtualmethod()
{ std::cout << "Derived nonvirtualmethod" << std::endl; }
virtual void virtualmethod()
{ std::cout << "Derived virtualmethod" << std::endl; }
};
int main()
{
Derived d;
Derived* pd = &d;
Base* pb = &d; //< NOTE: both pd and pb point to the same object
pd->nonvirtualmethod();
pb->nonvirtualmethod();
pd->virtualmethod();
pb->virtualmethod();
}
I gives you the following output:
Derived nonvirtualmethod
Base nonvirtualmethod
Derived virtualmethod
Derived virtualmethod //< invoked by a Base*
This is because there is a difference between the static type of the pb pointer (Base*)
and the dynamic type it points to (Derived).
The difference between virtual and plain methods is that non-virtual methods follow the the static type mapping (so a Base pointer invokes Base::methods), while virtual methods follow the chain of the runtime types, hence if a Base* points to a Derived, the Derived method will be called.
Destructors, in this sense, are nothing special: if it is not virtual, a Base pointer will not invoke the Derived one, hence you are left with an half-destroyed object, that is given back to the memory store.
The reason why this is UB (and not simply denied), is because the "memory store" is not managed by the language itself, but from the platform the program is hosted in: the crash most likely depends on the fact that the missing of the Derived part (still alive) will result in the operating system trying free a block of memory with a wrong starting address.