Using shared_from_this() without managed shared pointer in C++11 - c++

Let's say I have a class which is a child class of enable_shared_from_this. The documentation of this base class says there should be a shared pointer which owns this class before calling shared_from_this. Is it safe to allocate the class with new and call shared_from_this to manage the object?

As already mentioned by other users, calls to shared_from_this on instances that are not owned by shared_ptrs will result in an undefined behavior (usually an exception, but there are no guarantees).
So, why one more answer?
Because I did myself the same question once and got almost the same answer, then I started struggling with another question that immediately arose after that - how can I guarantee thus that all the instances are managed by a shared_ptr?
For the sake of completeness, I add another answer with a few details about this aspect.
Here a simple solution that had not been mentioned before.
So simple a solution, indeed: private constructors, factory method and variadic templates.
It follows a snippet that mixes all of them together in a minimal example:
#include<memory>
#include<utility>
class C: public std::enable_shared_from_this<C> {
C() = default;
C(const C &) = default;
C(C &&) = default;
C& operator=(const C &) = default;
C& operator=(C &&c) = default;
public:
template<typename... Args>
static std::shared_ptr<C> create(Args&&... args) noexcept {
return std::shared_ptr<C>{new C{std::forward<Args>(args)...}};
}
std::shared_ptr<C> ptr() noexcept {
return shared_from_this();
}
};
int main() {
std::shared_ptr<C> c1 = C::create();
std::shared_ptr<C> c2 = C::create(*c1);
std::shared_ptr<C> c3 = c2->ptr();
// these won't work anymore...
// C c4{};
// std::shared_ptr<C> c5 = std::make_shared<C>();
// std::shared_ptr<C> c6{new C{}};
// C c7{*c1};
// ... and so on ...
}
The basic (trivial?) idea is to forbid the explicit construction of new instances, but by using the factory method here called create.
Variadic templates are used to avoid writing several factory methods, nothing more. Perfect forwarding helps us to do that the right way.
Pretty simple, isn't it?
Anyway it took me a while to figure out that, so I hope this will help future readers once across the same doubt.

From the standard:
§ 20.8.2.4
shared_ptr shared_from_this();
shared_ptr shared_from_this() const;
7 *Requires: enable_shared_from_this shall be an accessible base class of T. this shall be a
subobject of an object t of type T. There shall be at least one shared_ptr instance p that owns &t.
8 Returns: A shared_ptr object r that shares ownership with p.
9 Postconditions: r.get() == this
If you call shared_from_this() within a class that is not managed by a shared_ptr the result will be undefined behaviour because you have not fulfilled one of the documented preconditions of the method.
I know from experience that in [the current version of] libc++ the result is an exception being thrown. However, like all undefined behavior this must not be relied upon.

The documentation of this base class says there should be a shared pointer which owns this [object] before calling shared_from_this.
Okay, cool.
Is it safe to allocate the [object] with new and call shared_from_this to manage the object?
No. There should be a shared pointer which owns this [object] before calling shared_from_this.

No, it's not safe. You should only call shared_from_this if the object is managed by a shared_ptr, not allocated via new (without an associated shared_ptr). For example this code
struct Test: std::enable_shared_from_this<Test> {
std::shared_ptr<Test> getptr() {
return shared_from_this();
}
};
Test *test = new Test;
std::shared_ptr<Test> test2 = test->getptr();
will throw std::bad_weak_ptr (at least when using libstdc++). But this is OK:
std::shared_ptr<Test> test(new Test);
std::shared_ptr<Test> test2 = test->getptr();

This is not safe at all.
Calling shared_from_this() from a non-shared_ptr invokes a bad_weak_ptr exception.
#include <iostream>
#include <memory>
struct A : std::enable_shared_from_this<A>
{
A(A *parent, int x): parent(parent), x(x) {
std::cout << "create A with " << x << std::endl;
}
void print(){
std::cout << "print A : " << x << std::endl;
}
~A(){
std::cout << "delete A" << std::endl;
}
A *parent;
int x;
};
void useA(const std::shared_ptr<A> &a) {
a->print();
a->parent->print();
for(auto parent = a->parent; parent; parent = parent->parent){
auto aptr = parent->shared_from_this();
aptr->print();
}
}
int main() {
auto a1 = new A(NULL, 0);
auto p1 = std::make_shared<A>(a1, 1);
std::cout << "main" << std::endl;
useA(p1);
}
In this example, std::bad_weak_ptr is thrown at
auto aptr = parent->shared_from_this();

Related

placement new to circumvent assignment constructor

Is it a viable solution to use placement new to circumvent copy assignment?
I have a member object that contains const members.
The object itself is meant to be created at runtime, but it's members are meant to be const.
I was wondering whether I could use placement new to circumvent the copy assignment at runtime?
#include <new>
#include <iostream>
struct A {
const int a;
A() : a(0) {};
A(int a) : a(a){}
};
struct B {
A a;
void createA() {
new(&a) A(69);
}
void useA() {
std::cout << a.a << std::endl;
}
};
int main(void) {
B b;
b.createA();
b.useA();
return 0;
}
It compiles and runs but does it invoke UB?
You should not do this, though it will technically work in this case.
By circumventing the copy assignment you are not giving the object a chance to perform any required cleanup on the old data (e.g. freeing memory resources). In this example, you are calling the new placement operator before the lifetime of a ends.
The following will work as it does not violate a's lifetime:
a.~A(); // manually destruct a, end it's lifetime
new(&a) A(69); // construct a new instance and start a new lifetime

Why do I get an exception for this usage of enable_shared_from_this()?

On the following example below, something is escaping me. Why does this generate the usual weak_ptr exception when using the shared_from_this, if there is an outstanding shared_ptr reference to the object which calls it ?
class A : std::enable_shared_from_this<A> {
public:
static std::shared_ptr<A> create() {
return std::shared_ptr<A>(new A());
}
A() {}
void setParent(const std::shared_ptr<A>& other) {}
std::shared_ptr<A> keep() {
auto o = A::create();
o->setParent(shared_from_this());
return o;
}
};
int main() {
std::shared_ptr<A> a = A::create();
auto s = a->keep();
}
You need to inherit publicly from enable_shared_from_this so the shared ptr ctor can see it.
As an aside, clang fails to compile your example due to this.
enable_shared_from_this must be an unambiguous and accessible base of A for the library to handle the internal weak pointer correctly.
Your inheritance is private (the default when using the class keyword).

Is it possible to change a C++ object's class after instantiation?

I have a bunch of classes which all inherit the same attributes from a common base class. The base class implements some virtual functions that work in general cases, whilst each subclass re-implements those virtual functions for a variety of special cases.
Here's the situation: I want the special-ness of these sub-classed objects to be expendable. Essentially, I would like to implement an expend() function which causes an object to lose its sub-class identity and revert to being a base-class instance with the general-case behaviours implemented in the base class.
I should note that the derived classes don't introduce any additional variables, so both the base and derived classes should be the same size in memory.
I'm open to destroying the old object and creating a new one, as long as I can create the new object at the same memory address, so existing pointers aren't broken.
The following attempt doesn't work, and produces some seemingly unexpected behaviour. What am I missing here?
#include <iostream>
class Base {
public:
virtual void whoami() {
std::cout << "I am Base\n";
}
};
class Derived : public Base {
public:
void whoami() {
std::cout << "I am Derived\n";
}
};
Base* object;
int main() {
object = new Derived; //assign a new Derived class instance
object->whoami(); //this prints "I am Derived"
Base baseObject;
*object = baseObject; //reassign existing object to a different type
object->whoami(); //but it *STILL* prints "I am Derived" (!)
return 0;
}
You can at the cost of breaking good practices and maintaining unsafe code. Other answers will provide you with nasty tricks to achieve this.
I dont like answers that just says "you should not do that", but I would like to suggest there probably is a better way to achieve the result you seek for.
The strategy pattern as suggested in a comment by #manni66 is a good one.
You should also think about data oriented design, since a class hierarchy does not look like a wise choice in your case.
Yes and no. A C++ class defines the type of a memory region that is an object. Once the memory region has been instantiated, its type is set. You can try to work around the type system sure, but the compiler won't let you get away with it. Sooner or later it will shoot you in the foot, because the compiler made an assumption about types that you violated, and there is no way to stop the compiler from making such assumption in a portable fashion.
However there is a design pattern for this: It's "State". You extract what changes into it's own class hierarchy, with its own base class, and you have your objects store a pointer to the abstract state base of this new hierarchy. You can then swap those to your hearts content.
No it's not possible to change the type of an object once instantiated.
*object = baseObject; doesn't change the type of object, it merely calls a compiler-generated assignment operator.
It would have been a different matter if you had written
object = new Base;
(remembering to call delete naturally; currently your code leaks an object).
C++11 onwards gives you the ability to move the resources from one object to another; see
http://en.cppreference.com/w/cpp/utility/move
I'm open to destroying the old object and creating a new one, as long as I can create the new object at the same memory address, so existing pointers aren't broken.
The C++ Standard explicitly addresses this idea in section 3.8 (Object Lifetime):
If, after the lifetime of an object has ended and before the storage which the object occupied is reused or released, a new object is created at the storage location which the original object occupied, a pointer that pointed to the original object, a reference that referred to the original object, or the name of the original object will automatically refer to the new object and, once the lifetime of the new object has started, can be used to manipulate the new object <snip>
Oh wow, this is exactly what you wanted. But I didn't show the whole rule. Here's the rest:
if:
the storage for the new object exactly overlays the storage location which the original object occupied, and
the new object is of the same type as the original object (ignoring the top-level cv-qualifiers), and
the type of the original object is not const-qualified, and, if a class type, does not contain any non-static data member whose type is const-qualified or a reference type, and
the original object was a most derived object (1.8) of type T and the new object is a most derived object of type T (that is, they are not base class subobjects).
So your idea has been thought of by the language committee and specifically made illegal, including the sneaky workaround that "I have a base class subobject of the right type, I'll just make a new object in its place" which the last bullet point stops in its tracks.
You can replace an object with an object of a different type as #RossRidge's answer shows. Or you can replace an object and keep using pointers that existed before the replacement. But you cannot do both together.
However, like the famous quote: "Any problem in computer science can be solved by adding a layer of indirection" and that is true here too.
Instead of your suggested method
Derived d;
Base* p = &d;
new (p) Base(); // makes p invalid! Plus problems when d's destructor is automatically called
You can do:
unique_ptr<Base> p = make_unique<Derived>();
p.reset(make_unique<Base>());
If you hide this pointer and slight-of-hand inside another class, you'll have the "design pattern" such as State or Strategy mentioned in other answers. But they all rely on one extra level of indirection.
I suggest you use the Strategy Pattern, e.g.
#include <iostream>
class IAnnouncer {
public:
virtual ~IAnnouncer() { }
virtual void whoami() = 0;
};
class AnnouncerA : public IAnnouncer {
public:
void whoami() override {
std::cout << "I am A\n";
}
};
class AnnouncerB : public IAnnouncer {
public:
void whoami() override {
std::cout << "I am B\n";
}
};
class Foo
{
public:
Foo(IAnnouncer *announcer) : announcer(announcer)
{
}
void run()
{
// Do stuff
if(nullptr != announcer)
{
announcer->whoami();
}
// Do other stuff
}
void expend(IAnnouncer* announcer)
{
this->announcer = announcer;
}
private:
IAnnouncer *announcer;
};
int main() {
AnnouncerA a;
Foo foo(&a);
foo.run();
// Ready to "expend"
AnnouncerB b;
foo.expend(&b);
foo.run();
return 0;
}
This is a very flexible pattern that has at least a few benefits over trying to deal with the issue through inheritance:
You can easily change the behavior of Foo later on by implementing a new Announcer
Your Announcers (and your Foos) are easily unit tested
You can reuse your Announcers elsewhere int he code
I suggest you have a look at the age-old "Composition vs. Inheritance" debate (cf. https://www.thoughtworks.com/insights/blog/composition-vs-inheritance-how-choose)
ps. You've leaked a Derived in your original post! Have a look at std::unique_ptr if it is available.
You can do what you're literally asking for with placement new and an explicit destructor call. Something like this:
#include <iostream>
#include <stdlib.h>
class Base {
public:
virtual void whoami() {
std::cout << "I am Base\n";
}
};
class Derived : public Base {
public:
void whoami() {
std::cout << "I am Derived\n";
}
};
union Both {
Base base;
Derived derived;
};
Base *object;
int
main() {
Both *tmp = (Both *) malloc(sizeof(Both));
object = new(&tmp->base) Base;
object->whoami();
Base baseObject;
tmp = (Both *) object;
tmp->base.Base::~Base();
new(&tmp->derived) Derived;
object->whoami();
return 0;
}
However as matb said, this really isn't a good design. I would recommend reconsidering what you're trying to do. Some of other answers here might also solve your problem, but I think anything along the idea of what you're asking for is going to be kludge. You should seriously consider designing your application so you can change the pointer when the type of the object changes.
You can by introducing a variable to the base class, so the memory footprint stays the same. By setting the flag you force calling the derived or the base class implementation.
#include <iostream>
class Base {
public:
Base() : m_useDerived(true)
{
}
void setUseDerived(bool value)
{
m_useDerived = value;
}
void whoami() {
m_useDerived ? whoamiImpl() : Base::whoamiImpl();
}
protected:
virtual void whoamiImpl() { std::cout << "I am Base\n"; }
private:
bool m_useDerived;
};
class Derived : public Base {
protected:
void whoamiImpl() {
std::cout << "I am Derived\n";
}
};
Base* object;
int main() {
object = new Derived; //assign a new Derived class instance
object->whoami(); //this prints "I am Derived"
object->setUseDerived(false);
object->whoami(); //should print "I am Base"
return 0;
}
In addition to other answers, you could use function pointers (or any wrapper on them, like std::function) to achieve the necessary bevahior:
void print_base(void) {
cout << "This is base" << endl;
}
void print_derived(void) {
cout << "This is derived" << endl;
}
class Base {
public:
void (*print)(void);
Base() {
print = print_base;
}
};
class Derived : public Base {
public:
Derived() {
print = print_derived;
}
};
int main() {
Base* b = new Derived();
b->print(); // prints "This is derived"
*b = Base();
b->print(); // prints "This is base"
return 0;
}
Also, such function pointers approach would allow you to change any of the functions of the objects in run-time, not limiting you to some already defined sets of members implemented in derived classes.
There is a simple error in your program. You assign the objects, but not the pointers:
int main() {
Base* object = new Derived; //assign a new Derived class instance
object->whoami(); //this prints "I am Derived"
Base baseObject;
Now you assign baseObject to *object which overwrites the Derived object with a Base object. However, this does work well because you are overwriting an object of type Derived with an object of type Base. The default assignment operator just assigns all members, which in this case does nothing. The object cannot change its type and still is a Derived objects afterwards. In general, this can leads to serious problems e.g. object slicing.
*object = baseObject; //reassign existing object to a different type
object->whoami(); //but it *STILL* prints "I am Derived" (!)
return 0;
}
If you instead just assign the pointer it will work as expected, but you just have two objects, one of type Derived and one Base, but I think you want some more dynamic behavior. It sounds like you could implement the specialness as a Decorator.
You have a base-class with some operation, and several derived classes that change/modify/extend the base-class behavior of that operation. Since it is based on composition it can be changed dynamically. The trick is to store a base-class reference in the Decorator instances and use that for all other functionality.
class Base {
public:
virtual void whoami() {
std::cout << "I am Base\n";
}
virtual void otherFunctionality() {}
};
class Derived1 : public Base {
public:
Derived1(Base* base): m_base(base) {}
virtual void whoami() override {
std::cout << "I am Derived\n";
// maybe even call the base-class implementation
// if you just want to add something
}
virtual void otherFunctionality() {
base->otherFunctionality();
}
private:
Base* m_base;
};
Base* object;
int main() {
Base baseObject;
object = new Derived(&baseObject); //assign a new Derived class instance
object->whoami(); //this prints "I am Derived"
// undecorate
delete object;
object = &baseObject;
object->whoami();
return 0;
}
There are alternative patterns like Strategy which implement different use cases resp. solve different problems. It would probably good to read the pattern documentation with special focus to the Intent and Motivation sections.
I would consider regularizing your type.
class Base {
public:
virtual void whoami() { std::cout << "Base\n"; }
std::unique_ptr<Base> clone() const {
return std::make_unique<Base>(*this);
}
virtual ~Base() {}
};
class Derived: public Base {
virtual void whoami() overload {
std::cout << "Derived\n";
};
std::unique_ptr<Base> clone() const override {
return std::make_unique<Derived>(*this);
}
public:
~Derived() {}
};
struct Base_Value {
private:
std::unique_ptr<Base> pImpl;
public:
void whoami () {
pImpl->whoami();
}
template<class T, class...Args>
void emplace( Args&&...args ) {
pImpl = std::make_unique<T>(std::forward<Args>(args)...);
}
Base_Value()=default;
Base_Value(Base_Value&&)=default;
Base_Value& operator=(Base_Value&&)=default;
Base_Value(Base_Value const&o) {
if (o.pImpl) pImpl = o.pImpl->clone();
}
Base_Value& operator=(Base_Value&& o) {
auto tmp = std::move(o);
swap( pImpl, tmp.pImpl );
return *this;
}
};
Now a Base_Value is semantically a value-type that behaves polymorphically.
Base_Value object;
object.emplace<Derived>();
object.whoami();
object.emplace<Base>();
object.whoami();
You could wrap a Base_Value instance in a smart pointer, but I wouldn't bother.
I don’t disagree with the advice that this isn’t a great design, but another safe way to do it is with a union that can hold any of the classes you want to switch between, since the standard guarantees it can safely hold any of them. Here’s a version that encapsulates all the details inside the union itself:
#include <cassert>
#include <cstdlib>
#include <iostream>
#include <new>
#include <typeinfo>
class Base {
public:
virtual void whoami() {
std::cout << "I am Base\n";
}
virtual ~Base() {} // Every base class with child classes that might be deleted through a pointer to the
// base must have a virtual destructor!
};
class Derived : public Base {
public:
void whoami() {
std::cout << "I am Derived\n";
}
// At most one member of any union may have a default member initializer in C++11, so:
Derived(bool) : Base() {}
};
union BorD {
Base b;
Derived d; // Initialize one member.
BorD(void) : b() {} // These defaults are not used here.
BorD( const BorD& ) : b() {} // No per-instance data to worry about!
// Otherwise, this could get complicated.
BorD& operator= (const BorD& x) // Boilerplate:
{
if ( this != &x ) {
this->~BorD();
new(this) BorD(x);
}
return *this;
}
BorD( const Derived& x ) : d(x) {} // The constructor we use.
// To destroy, be sure to call the base class’ virtual destructor,
// which works so long as every member derives from Base.
~BorD(void) { dynamic_cast<Base*>(&this->b)->~Base(); }
Base& toBase(void)
{ // Sets the active member to b.
Base* const p = dynamic_cast<Base*>(&b);
assert(p); // The dynamic_cast cannot currently fail, but check anyway.
if ( typeid(*p) != typeid(Base) ) {
p->~Base(); // Call the virtual destructor.
new(&b) Base; // Call the constructor.
}
return b;
}
};
int main(void)
{
BorD u(Derived{false});
Base& reference = u.d; // By the standard, u, u.b and u.d have the same address.
reference.whoami(); // Should say derived.
u.toBase();
reference.whoami(); // Should say base.
return EXIT_SUCCESS;
}
A simpler way to get what you want is probably to keep a container of Base * and replace the items individually as needed with new and delete. (Still remember to declare your destructor virtual! That’s important with polymorphic classes, so you call the right destructor for that instance, not the base class’ destructor.) This might save you some extra bytes on instances of the smaller classes. You would need to play around with smart pointers to get safe automatic deletion, though. One advantage of unions over smart pointers to dynamic memory is that you don’t have to allocate or free any more objects on the heap, but can just re-use the memory you have.
DISCLAIMER: The code here is provided as means to understand an idea, not to be implemented in production.
You're using inheritance. It can achieve 3 things:
Add fields
Add methods
replace virtual methods
Out of all those features, you're using only the last one. This means that you're not actually forced to rely on inheritance. You can get the same results by many other means. The simplest is to keep tabs on the "type" by yourself - this will allow you to change it on the fly:
#include <stdexcept>
enum MyType { BASE, DERIVED };
class Any {
private:
enum MyType type;
public:
void whoami() {
switch(type){
case BASE:
std::cout << "I am Base\n";
return;
case DERIVED:
std::cout << "I am Derived\n";
return;
}
throw std::runtime_error( "undefined type" );
}
void changeType(MyType newType){
//insert some checks if that kind of transition is legal
type = newType;
}
Any(MyType initialType){
type = initialType;
}
};
Without inheritance the "type" is yours to do whatever you want. You can changeType at any time it suits you. With that power also comes responsibility: the compiler will no longer make sure the type is correct or even set at all. You have to ensure it or you'll get hard to debug runtime errors.
You may wrap it in inheritance just as well, eg. to get a drop-in replacement for existing code:
class Base : Any {
public:
Base() : Any(BASE) {}
};
class Derived : public Any {
public:
Derived() : Any(DERIVED) {}
};
OR (slightly uglier):
class Derived : public Base {
public:
Derived : Base() {
changeType(DERIVED)
}
};
This solution is easy to implement and easy to understand. But with more options in the switch and more code in each path it gets very messy. So the very first step is to refactor the actual code out of the switch and into self-contained functions. Where better to keep than other than Derivied class?
class Base {
public:
static whoami(Any* This){
std::cout << "I am Base\n";
}
};
class Derived {
public:
static whoami(Any* This){
std::cout << "I am Derived\n";
}
};
/*you know where it goes*/
switch(type){
case BASE:
Base:whoami(this);
return;
case DERIVED:
Derived:whoami(this);
return;
}
Then you can replace the switch with an external class that implements it via virtual inheritance and TADA! We've reinvented the Strategy Pattern, as others have said in the first place : )
The bottom line is: whatever you do, you're not inheriting the main class.
you cannot change to the type of an object after instantiation, as you can see in your example you have a pointer to a Base class (of type base class) so this type is stuck to it until the end.
the base pointer can point to upper or down object doesn't mean changed its type:
Base* ptrBase; // pointer to base class (type)
ptrBase = new Derived; // pointer of type base class `points to an object of derived class`
Base theBase;
ptrBase = &theBase; // not *ptrBase = theDerived: Base of type Base class points to base Object.
pointers are much strong, flexible, powerful as much dangerous so you should handle them cautiously.
in your example I can write:
Base* object; // pointer to base class just declared to point to garbage
Base bObject; // object of class Base
*object = bObject; // as you did in your code
above it's a disaster assigning value to un-allocated pointer. the program will crash.
in your example you escaped the crash through the memory which was allocated at first:
object = new Derived;
it's never good idea to assign a value and not address of a subclass object to base class. however in built-in you can but consider this example:
int* pInt = NULL;
int* ptrC = new int[1];
ptrC[0] = 1;
pInt = ptrC;
for(int i = 0; i < 1; i++)
cout << pInt[i] << ", ";
cout << endl;
int* ptrD = new int[3];
ptrD[0] = 5;
ptrD[1] = 7;
ptrD[2] = 77;
*pInt = *ptrD; // copying values of ptrD to a pointer which point to an array of only one element!
// the correct way:
// pInt = ptrD;
for(int i = 0; i < 3; i++)
cout << pInt[i] << ", ";
cout << endl;
so the result as not as you guess.
I have 2 solutions. A simpler one that doesn't preserve the memory address, and one that does preserve the memory address.
Both require that you provide provide downcasts from Base to Derived which isn't a problem in your case.
struct Base {
int a;
Base(int a) : a{a} {};
virtual ~Base() = default;
virtual auto foo() -> void { cout << "Base " << a << endl; }
};
struct D1 : Base {
using Base::Base;
D1(Base b) : Base{b.a} {};
auto foo() -> void override { cout << "D1 " << a << endl; }
};
struct D2 : Base {
using Base::Base;
D2(Base b) : Base{b.a} {};
auto foo() -> void override { cout << "D2 " << a << endl; }
};
For the former one you can create a smart pointer that can seemingly change the held data between Derived (and base) classes:
template <class B> struct Morpher {
std::unique_ptr<B> obj;
template <class D> auto morph() {
obj = std::make_unique<D>(*obj);
}
auto operator->() -> B* { return obj.get(); }
};
int main() {
Morpher<Base> m{std::make_unique<D1>(24)};
m->foo(); // D1 24
m.morph<D2>();
m->foo(); // D2 24
}
The magic is in
m.morph<D2>();
which changes the held object preserving the data members (actually uses the cast ctor).
If you need to preserve the memory location, you can adapt the above to use a buffer and placement new instead of unique_ptr. It is a little more work a whole lot more attention to pay to, but it gives you exactly what you need:
template <class B> struct Morpher {
std::aligned_storage_t<sizeof(B)> buffer_;
B *obj_;
template <class D>
Morpher(const D &new_obj)
: obj_{new (&buffer_) D{new_obj}} {
static_assert(std::is_base_of<B, D>::value && sizeof(D) == sizeof(B) &&
alignof(D) == alignof(B));
}
Morpher(const Morpher &) = delete;
auto operator=(const Morpher &) = delete;
~Morpher() { obj_->~B(); }
template <class D> auto morph() {
static_assert(std::is_base_of<B, D>::value && sizeof(D) == sizeof(B) &&
alignof(D) == alignof(B));
obj_->~B();
obj_ = new (&buffer_) D{*obj_};
}
auto operator-> () -> B * { return obj_; }
};
int main() {
Morpher<Base> m{D1{24}};
m->foo(); // D1 24
m.morph<D2>();
m->foo(); // D2 24
m.morph<Base>();
m->foo(); // Base 24
}
This is of course the absolute bare bone. You can add move ctor, dereference operator etc.
#include <iostream>
class Base {
public:
virtual void whoami() {
std::cout << "I am Base\n";
}
};
class Derived : public Base {
public:
void whoami() {
std::cout << "I am Derived\n";
}
};
Base* object;
int main() {
object = new Derived;
object->whoami();
Base baseObject;
object = &baseObject;// this is how you change.
object->whoami();
return 0;
}
output:
I am Derived
I am Base
Your assignment only assigns member variables, not the pointer used for virtual member function calls. You can easily replace that with full memory copy:
//*object = baseObject; //this assignment was wrong
memcpy(object, &baseObject, sizeof(baseObject));
Note that much like your attempted assignment, this would replace member variables in *object with those of the newly constructed baseObject - probably not what you actually want, so you'll have to copy the original member variables to the new baseObject first, using either assignment operator or copy constructor before the memcpy, i.e.
Base baseObject = *object;
It is possible to copy just the virtual functions table pointer but that would rely on internal knowledge about how the compiler stores it so is not recommended.
If keeping the object at the same memory address is not crucial, a simpler and so better approach would be the opposite - construct a new base object and copy the original object's member variables over - i.e. use a copy constructor.
object = new Base(*object);
But you'll also have to delete the original object, so the above one-liner won't be enough - you need to remember the original pointer in another variable in order to delete it, etc. If you have multiple references to that original object you'll need to update them all, and sometimes this can be quite complicated. Then the memcpy way is better.
If some of the member variables themselves are pointers to objects that are created/deleted in the main object's constructor/destructor, or if they have a more specialized assignment operator or other custom logic, you'll have some more work on your hands, but for trivial member variables this should be good enough.

Can upcasting a shared_ptr<T> to a shared_ptr<void> lead to undefined behaviour?

Shared pointers are quite smart. They remember the type they where first constructed with in order to delete them correctly. Take that for example:
struct A { virtual void test() = 0; };
struct B : A { void test() override {} };
void someFunc() {
std::shared_ptr<A> ptr1;
ptr1 = std::make_shared<B>();
// Here at the end of the scope, B is deleted correctly
}
However, there seems to be a problem with void pointers: for a downcast of a void pointer to be valid, one must downcast it to the type it was originally upcasted from.
For example:
void* myB = new B;
// Okay, well defined
doStuff(static_cast<B*>(myB));
// uh oh, not good!
// For the same instance of a child object, a pointer to the base and
// a pointer to the child can be differrent.
doStuff(static_cast<A*>(myB));
With std::shared_ptr, when you use std::make_shared, the deleter must look similar to this function: [](B* ptr){ delete ptr; }. Since the pointer (in the first example) is holding a B instance in a pointer to A and deletes it correctly, it must downcast it in some way.
My question is: is the following code snippet invokes undefined behaviour?
void someFunc() {
{
std::shared_ptr<void> ptr = std::make_shared<B>();
// Deleting the pointer seems okay to me,
// the shared_ptr knows that a B was originally allocated with a B and
// will send the void pointer to the deleter that's delete a B.
}
std::shared_ptr<void> vptr;
{
std::shared_ptr<A> ptr = std::make_shared<B>();
// ptr is pointing to the base, which can be
// different address than the pointer to the child.
// assigning the pointer to the base to the void pointer.
// according to my current knowledge of void pointers,
// any future use of the pointer must cast it to a A* or end up in UB.
vptr = ptr;
}
// is the pointer deleted correctly or it tries to
// cast the void pointer to a B pointer without downcasting to a A* first?
// Or is it well defined and std::shared_ptr uses some dark magic unknown to me?
}
The code is correct.
std::shared_ptr internally saves the real pointer and the real deleter as they are in the constructor, so no matter how you downcast it, as long as the downcast is valid the deleter will be right.
The shared_ptr actually does not hold a pointer to the object, but a pointer to an intermediate struct that holds the actual object, the reference counter and the deleter. It doesn't matter if you cast the shared_ptr, that intermediate struct does not change. It cannot change because your vptr and ptr, although of different types, share the reference counter (and the object and the deleter, of course).
BTW, that intermediate struct is the reason for the make_shared optimization: it allocates both the intermediate struct and the object itself in the same memory block and avoids the extra allocation.
To illustrate how smart pointers can be, I have written a program with plain pointers that crashes (with GCC 6.2.1) because of your problem:
#include <memory>
#include <iostream>
struct A
{
int a;
A() :a(1) {}
~A()
{
std::cout << "~A " << a << std::endl;
}
};
struct X
{
int x;
X() :x(3) {}
~X()
{
std::cout << "~X " << x << std::endl;
}
};
struct B : X, A
{
int b;
B() : b(2) {}
~B()
{
std::cout << "~B " << b << std::endl;
}
};
int main()
{
A* a = new B;
void * v = a;
delete (B*)v; //crash!
return 0;
}
Actually it prints the wrong integer values, which proves the UB.
~B 0
~A 2
~X 1
*** Error in `./a.out': free(): invalid pointer: 0x0000000001629c24 ***
But the version with smart pointers works just fine:
int main()
{
std::shared_ptr<void> vptr;
{
std::shared_ptr<A> ptr = std::make_shared<B>();
vptr = ptr;
}
return 0;
}
It prints as expected:
~B 2
~A 1
~X 3
The shared_ptr always passes the original pointer to the deleter, not the one obtained through vptr.get(). This is necessary not only to make this case work, but also to have pointers to member subobjects and owned objects as embodied in the constructor overload shared_ptr<T>::shared_ptr(const shared_ptr<T>&, element_type*).
So this is safe.

Does C++ have existing method to check if an object is a derived-typed object?

Does C++ have existing method to check if an object is a derived-typed object? For example,
class A
{};
class B : public A
{};
A* p;
And check if p points to B.
If the class is polymorphic (i.e., has at least one virtual member function), you can use dynamic_cast or typeid.
Otherwise, no. Keeping track of an object's dynamic type has a cost, and the language was designed to avoid pessimizing code that has no need for it.
And check if p points to B.
You can use dynamic_cast for that, if there is at least one virtual member function in the class. It is common to make the destructor virtual.
class A
{
virtual ~A() {}
};
and then,
B* bPtr = dynamic_cast<B*>(p);
if ( bPtr )
{
// Use the pointer
}
Does C++ have existing method to check if an object is a derived-typed object?
There are actually two ways to achieve this:
A* p = new B();
B* pB = static_cast<B*>(p); // Checks if B is related to A in an inheritance
// ^^^^^^^^^^^^^^^^^^^ hierarchy. Fails to compile if not.
A* pA = new B();
B* pB = dynamic_cast<B*>(pA); // Checks if pA actually points to an instance of B
// ^^^^^^^^^^^^^^^^^^^^ at runtime, and returns nullptr if not
if(pB) {
// do stuff with B
}
The latter example requires you have a virtual base class:
class A {
public:
virtual ~A() {} // <<<<<<<<<<<<<<<<<<
};
If the object in question is not a polymorphic class it is possible to determine whether a class object pointed to has a specific base class at compile time. This sometimes occurs in template code where different base classes are possible.
You use std::is_base_of as follows: Note that you must also use std::remove_reference since *p is an lvalue and decltype() produces a reference.
#include <type_traits>
#include <iostream>
class A {};
class B : public A{};
int main() {
A a;
B b;
A* pa=&a;
B* pb=&b;
std::cout << std::is_base_of<A, B>::value << "\n"; // true
std::cout << std::is_base_of<B, A>::value << "\n"; // false
std::cout << std::is_base_of<A, std::remove_reference<decltype(*pb)>::type>::value << "\n"; // true
std::cout << std::is_base_of<B, std::remove_reference<decltype(*pa)>::type>::value << "\n"; // false
}