What is the difference between QPointer, QSharedPointer and QWeakPointer classes in Qt? - c++

I have read from the Qt documentations about QPointer, QSharedPointer and QWeakPointer classes. It says:
QPointer is a template class that provides guarded pointers to Qt objects and behaves like a normal C++ pointer except that it is automatically set to 0 when the referenced object is destroyed and no "dangling pointers" are produced.
QSharedPointer class holds a strong reference to a shared pointer.
QWeakPointer class holds a weak reference to a shared pointer.
My questions is "What is the difference between these classes?". i.e what is the difference between a pointer to an object and a reference to a pointer? Are they all pointers to objects with different mechanisms and behaviors?

QPointer:
QPointer can only point to QObject instances. It will be automatically set to nullptr if the pointed to object is destroyed. It is a weak pointer specialized for QObject.
Consider this fragment:
QObject *obj = new QObject;
QPointer<QObject> pObj(obj);
delete obj;
Q_ASSERT(pObj.isNull()); // pObj will be nullptr now
QSharedPointer
A reference-counted pointer. The actual object will only be deleted, when all shared pointers are destroyed. Equivalent to std::shared_ptr.
int *pI = new int;
QSharedPointer<int> pI1(pI);
QSharedPointer<int> pI2 = pI1;
pI1.clear();
// pI2 is still pointing to pI, so it is not deleted
pI2.clear();
// No shared pointers anymore, pI is deleted
Note that as long as there is a shared pointer, the object is not deleted!
QWeakPointer:
Can hold a weak reference to a shared pointer. It will not prevent the object from being destroyed, and is simply reset. Equivalent to std::weak_ptr, where lock is equivalent to toStrongRef.
int *pI = new int;
QSharedPointer<int> pI1(pI);
QWeakPointer<int> pI2 = pI1;
pI1.clear();
// No shared pointers anymore, pI is deleted
//
// To use the shared pointer, we must "lock" it for use:
QSharedPointer<int> pI2_locked = pI2.toStrongRef();
Q_ASSERT(pI2_locked.isNull());
This can be used if you need access to an object that is controlled by another module.
To use a weak pointer, you must convert it to a QSharedPointer. You should never base a decision on the weak pointer being valid. You can only use data() or isNull() to determine that the pointer is null.
Generally, to use a weak pointer, you must convert it to a shared pointer since such an operation ensures that the object will survive for as long as you are using it. This is equivalent to "locking" the object for access and is the only correct way of using the object pointed to by a weak pointer.
QScopedPointer:
This is just a helper class that will delete the referenced object when the pointer goes out of scope. Thus, binds a dynamically allocated object to a variable scope.
You can use this for RAII semantics for locals, e.g.:
MyClass *foo() {
QScopedPointer<MyClass> myItem(new MyClass);
// Some logic
if (some condition) {
return nullptr; // myItem will be deleted here
}
return myItem.take(); // Release item from scoped pointer and return it
}
The item will also be deleted in case of an exception
Another use case can be member variables of an object. Then you don't need to write a destructor for those:
class MyClass {
public:
MyClass() : myPtr(new int) {}
private:
QScopedPointer<int> myPtr; // Will be deleted automatically when containing object is deleted
}

QSharedPointer : std::shared_ptr
QWeakPointer : std::weak_ptr
QScopedPointer : std::unique_ptr
QPointer : no STL equivalent. Nulled when the QObject destructs.

Related

Does QPointer::clear() delete its referenced pointer, or does "Clears this QPointer object." mean something else?

QPointer has a method, clear().
Clears this QPointer object.
I am not sure what "clear" exactly means. In my mind, it could mean
It deletes the pointer you referenced.
or
It un-attaches the pointer you referenced, leaving that pointer on the heap, and the QPointer<T> object no longer tied to any pointer.
Maybe it means something else? Could you please let me know what it actually does?
QPointer is a tracking pointer. It tracks the lifetime of an object. It doesn't do any owning duties. It never will deallocate any storage owned by QObject. It can deallocate the underlying implementation detail - the shared reference object, but that doesn't affect anything really that the user cares about; those objects are deallocated only when the underlying QObject is gone and the last QPointer is being destructed.
It deletes the pointer you referenced.
That's IMHO a rather confusing language. Pointers are values. To reference a pointer has a well established meaning:
const int *q = ....; // a pointer
*q; // a pointer dereference (not used for anything)
To me, "deleting a pointer" is this:
// dynamically allocate a pointer
int **p = new int*();
// make it point somewhere
int i = {};
assert(0 == i);
*p = &i;
// use it
**p = 44;
assert(44 == i);
// delete it
delete p; // a pointer-to-integer got deleted
It un-attaches the pointer you referenced, leaving that pointer on the heap, and the QPointer<T> object no longer tied to any pointer.
It's pointers galore for sure, but that's veritable goobledygook. Just because I might understand what you mean doesn't imply that anyone should talk that way :)
A QPointer<T> tracks the lifetime of a T-instance, an object. It's objects it tracks, not pointers. So clear() makes the QPointer not track whatever object of type T it was tracking. That's all. And that's how to say it without making everyone doubt their sanity :)
It is true that the way you make a QPointer track an object is by pointing to it via a raw pointer. That's just how you get a QPointer going, that's all.
It is incorrect to conflate QPointer with heap - no heap is involved in the example below, at least not explicitly. The obj instance is an automatic variable. Implementations are free to put it on the dynamic store of some kind - even a literal heap, but that's typical of C++ interpreters and not what we're usually used to :)
#include <QtCore>
int main() {
QPointer<QObject> p;
Q_ASSERT(p.isNull());
{
QObject obj;
p = &obj;
Q_ASSERT(!p.isNull());
}
Q_ASSERT(p.isNull());
}

boost::shared_ptr from pointer

I just stumbled on the boost::shared_ptr documentation, which goes:
Sometimes it is necessary to obtain a shared_ptr given a raw pointer
to an object that is already managed by another shared_ptr instance.
Example:
void f(X * p)
{
shared_ptr<X> px(???);
}
Inside f, we'd like to create a shared_ptr to *p.
In the general case, this problem has no solution.
Why? Is it not allowed to do something like:
shared_ptr<X> px(p);
Am I missing something?
If you have a shared_ptr managing a pointer and then create another shared_ptr managing the same pointer (NOT copying the original shared_ptr), you end up with two managers for the same resource. When one of the two reach a reference count of 0, it will delete the object and the other shared_ptr will point to deleted memory with all that follows.
If you would do this:
main() {
Object *obj = new Object();
func(obj)
}
void func( Object *obj ) {
shared_ptr objPtr(obj); // take ownership.
objPtr->fun();
// Passed object "obj" is destroyed here.
}
At the end of the function func the object pointer would get destroyed, and with the pointer the object itself. This wouldn't be a desirable behaviour.
Actually, you can do that, but you must know that the pointed object will be deleted when exiting the function...
I tried with boost:
void f(X * p)
{
boost::shared_ptr<X> px(p);
// do smething
}
void main()
{
X* ptr = new X();
f( ptr );
// ptr is not valid anymore because the object were deleted
}
Jean
You could do it, but it could lead to undefined behavior, since there is no way to tell the second shared pointer that the reference count (the number of shared pointers pointing at the same object) increased. Then things like this could happen:
void f()
{
boost::shared_ptr<int> firstSmart(new int(23)); // firstSmart is the only
// manager of the int
int *raw = firstSmart.get();
boost::shared_ptr<int> secondSmart(raw); // secondSmart also manages
// the same int as firstSmart
// but secondSmart does not
// know about firstSmart
// and vice versa
}
when f exits secondSmart gets destroyed, destroying the shared int. Then firstSmart gets destroyed and attempts to destroy the already destroyed int thus leading to undefined behaviour.
Its not possible to do this.
Shared pointers work using reference counting. When you assign a resource (raw pointer) to a shared pointer a reference count object is created with count =1. When another shared pointer is created for the same resource the reference count object (which is shared between both the shared pointers) is updated to value count =2. If one shared pointer is deleted the count in the shared reference object is decremented and when it reached 0 the resource is destroyed.
For the above mechanism to work the first shared pointer should be created using something like shared_ptr px(p) and all subsequent ones using px (not 'p'). This way all the shared pointers created will know that they are holding same resource and share the same reference count object.
If you created another shared pointer using shared_ptr px(p) then you end up with two shared pointer not related to each other - i.e there reference count objects are not same. They both assume that they are holding distinct resource and each of them have distinct (different) reference count object with count =1. (You don’t want this).

Should a pointer be the same before and after adding to a unique_ptr?

I have a std::vector of unique_ptrs and I'm happy to have them manage the life cycle of those objects.
However I require storing other pointers to those objects for convenience. I know that once unique_ptr removes something, those other pointers will dangle. But I'm more concerned about the validity of those pointers before and after unique_ptr gets them.
I do not always create via new within the unique_ptr itself, for example I might pass new Something as a function parameter in which case the unique_ptr is using move on that pointer into itself inside the function.
But I might also new Something before I pass it into a function that then assigned it a unique_ptr.
Once an object is assigned to a unique_ptr I can get a pointer to it via get(). But can I always assume that this get() pointer points to the same place as the pointer initially obtained via new if the original pointer was created before the assignment to a unique_ptr ?
My assumption is Yes, and that even if the vector resizes and reallocates, the unique_ptr as well as any other pointers to the objects in memory remain the same.
Yes, a std::unique_ptr<T> holds a pointer to T, and it will not alter the value between initialization and later retrieval with get()
A common use of a unique_ptr is to assign one "parent" object ownership of a dynamically-allocated "subobject", in a similiar same way as:
struct A
{
B b;
}
int main()
{
A a = ...;
B* p = &a.b;
}
In the above b is a true subobject of A.
Compare this to:
struct A
{
unique_ptr<B> b = new B(...);
}
int main()
{
A a = ...;
B* p = a.b.get();
}
In the above A and (*b) have a similar relationship to the first example, except here the B object is allocated on the heap. In both cases the destructor of A will destroy the "subobject". This "on heap" subobject structure may be preferable in some cases, for example because B is a polymorphic base type, or to make B an optional/nullable subobject of A.
The advantage of using unique_ptr over a raw pointer to manage this ownership relationship is that it will automatically destroy it in As destructor, and it will automatically move construct and move assign it as part of A.
As usual, in both cases, you must be careful that the lifetime of any raw pointers to the subobject are enclosed by the lifetime of the owning object.
Yes, you are correct, because unique_ptr does not copy the object; therefore, it has to point to the same address. However, once you give a pointer to a unique_ptr to own, you should not use that raw pointer any more, because the unique_ptr could be destroyed and deallocate the memory, and turn your raw pointer into a dangling pointer. Perhaps shared_ptr would be better for your situation.

how to point to a shared_ptr

So I have a shared_ptr in my Main class, and I'd like some other object (a singleton class) to have access to what the shared_ptr is pointing to.
In pseudo code...
mySingletonInstance->somePointer = myShared_ptr;
How do I do that?
If the singleton should participate in shared management of the object that is held by shared_ptr then it's somePointer could be simply a copy of that shared_ptr. Otherwise use a weak_ptr.
If you don't want mySingletonInstance->somePointer to respect the ownership semantics of shared_ptr, then:
class MySingletonClass {
SomeType* somePointer;
}
shared_ptr<SomeType> myShared_ptr;
...
// Then either of the following lines:
mySingletonInstance->somePointer = mySharedPtr.get();
mySingletonInstance->somePointer = &*mySharedPtr;
But beware -- your somePointer might hold a pointer to the object even after it has been deleted.

shared_ptr deletes the object

void ClassName::LocalMethod( )
{
boost::shared_ptr<ClassName> classNamePtr( this );
//some operation with classNamePtr
return;
}
Here the object is getting released when it returns from LocalMethod() since classNamePtr is out of scope. Isn't the shared_ptr smart enough to know that the ClassName object is still in scope and not to delete it?
What does it mean to create a shared_ptr to an object? It means that the holder of the shared_ptr now assumes ownership over the object. Ownership meaning that the object will be deleted when he so desires. When the holder of the shared_ptr destroys its shared_ptr, that will cause the object to potentially be destroyed, assuming that there are no other shared_ptrs to that object.
When a shared_ptr is a member of a class, that means that the lifetime of the object pointed to by the shared_ptr is at least as long as the object that the shared_ptr is a member of. When a shared_ptr is on the stack, this means that the lifetime of the object that the shared_ptr is pointing to will be at least as long as the scope it was created in. Once the object falls off the stack, it may be deleted.
The only time you should ever take a pointer and wrap it into a shared_ptr is when you are allocating the object initially. Why? Because an object does not know whether it is in a shared_ptr or not. It can't know. This means that the person who creates the original shared_ptr now has the responsibility to pass it around to other people who need to share ownership of that memory. The only way shared ownership works is through the copy constructor of shared_ptr. For example:
shared_ptr<int> p1 = new int(12);
shared_ptr<int> p2 = p1.get();
shared_ptr<int> p3 = p1;
The copy constructor of shared_ptr creates shared ownership between p1 and p3. Note that p2 does not share ownership with p1. They both think they have ownership over the same memory, but that's not the same as sharing it. Because they both think that they have unique ownership of it.
Therefore, when the three pointers are destroyed, the following will happen. First, p3 will be destroyed. But since p3 and p1 share ownership of the integer, the integer will not be destroyed yet. Next, p2 will be destroyed. Since it thinks that it is the only holder of the integer, it will then destroy it.
At this point, p1 is pointing to deleted memory. When p1 is destroyed, it thinks that it is the only holder of the integer, so it will then destroy it. This is bad, since it was already destroyed.
Your problem is this. You are inside an instance of a class. And you need to call some functions of yours that take a shared_ptr. But all you have is this, which is a regular pointer. What do you do?
You're going to get some examples that suggest enable_shared_from_this. But consider a more relevant question: "why do those functions take a shared_ptr as an argument?"
The type of pointer a function takes is indicative of what that function does with its argument. If a function takes a shared_ptr, that means that it needs to own the pointer. It needs to take shared ownership of the memory. So, look at your code and ask whether those functions truly need to take ownership of the memory. Are they storing the shared_ptr somewhere long-term (ie: in an object), or are they just using them for the duration of the function call?
If it's the latter, then the functions should take a naked pointer, not a shared_ptr. That way, they cannot claim ownership. Your interface is then self-documenting: the pointer type explains ownership.
However, it is possible that you could be calling functions that truly do need to take shared ownership. Then you need to use enable_shared_from_this. First, your class needs to be derived from enable_shared_from_this. Then, in the function:
void ClassName::LocalMethod()
{
boost::shared_ptr<ClassName> classNamePtr(shared_from_this());
//some operation with classNamePtr
return;
}
Note that there is a cost here. enable_shared_from_this puts a boost::weak_ptr in the class. But there is no virtual overhead or somesuch; it doesn't make the class virtual. enable_shared_from_this is a template, so you have to declare it like this:
class ClassName : public boost::enable_shared_from_this<ClassName>
Isn't the shared_ptr smart enough to know that the ClassName object is
still in scope and not to delete it?
That's not how shared_ptr works. When you pass a pointer while constructing a shared_ptr, the shared_ptr will assume ownership of the pointee (in this case, *this). In other words, the shared_ptr assumes total control over the lifetime of the pointee by virtue of the fact that the shared_ptr now owns it. Because of this, the last shared_ptr owning the pointee will delete it.
If there will be no copies of classNamePtr outside of ClassName::LocalMethod(), you can pass a deleter that does nothing while constructing classNamePtr. Here's an example of a custom deleter being used to prevent a shared_ptr from deleting its pointee. Adapting the example to your situation:
struct null_deleter // Does nothing
{
void operator()(void const*) const {}
};
void ClassName::LocalMethod()
{
// Construct a shared_ptr to this, but make it so that it doesn't
// delete the pointee.
boost::shared_ptr<ClassName> classNamePtr(this, null_deleter());
// Some operation with classNamePtr
// The only shared_ptr here will go away as the stack unwinds,
// but because of the null deleter it won't delete this.
return;
}
You can also use enable_shared_from_this to obtain a shared_ptr from this. Note that the member function shared_from_this() only works if you have an existing shared_ptr already pointing to this.
class ClassName : public enable_shared_from_this<ClassName>
{
public:
void LocalMethod()
{
boost::shared_ptr<ClassName> classNamePtr = shared_from_this();
}
}
// ...
// This must have been declared somewhere...
shared_ptr<ClassName> p(new ClassName);
// before you call this:
p->LocalMethod();
This is the more appropriate, "official" method and it's much less hackish than the null deleter method.
It could also be that you don't actually need to create a shared_ptr in the first place. What goes into the section commented //some operation with classNamePtr? There might be an even better way than the first two ways.