I see below notes in std::unique_ptr reference:
Only non-const unique_ptr can transfer the ownership of the managed object to another unique_ptr. The lifetime of an object managed by const std::unique_ptr is limited to the scope in which the pointer was created.
Is there anyone who can explain it with an example? I could not figure it out why.
You simply can not move from a const std::unique_ptr and you can't use other modifying member functions - swap, release and reset either (these are logically non-const qualified, cannot be called on a const instance).
Transferring ownership implies resetting the old owner to non-owning state, thus modifying it.
const std::unique_ptr will manage at most one object during its lifetime (starting from its initialization).In case of std::unique_ptr const&, you won't be able to transfer ownership from the referenced (even non-const) std::unique_ptr through this reference (const correctness).
The reset, release, swap, and move assignment functions are non-const member functions and therefore cannot be used with a const instance of the std::unique_ptr class. Therefore, a const std::unique_ptr has no way of being modified and is forced to own the pointer until it goes out of scope.
Normally you can transfer ownership of a managed object to another unique_ptr by using move assignment or the move constructor, e.g.:
std::unique_ptr<int> p(new int(1));
std::unique_ptr<int> q(std::move(p));
//now q owns the pointer, which is freed when q is destructed
But if p was const, you wouldn't be able to do so, and the managed object will be freed when p is destructed:
const std::unique_ptr<int> p(new int(1));
std::unique_ptr<int> q(std::move(p)); //compilation error
A unique_ptr owns the memory that it points at.
void MyFunc()
{
// Creates a unique pointer that points at a Foo object.
std::unique_ptr<Foo> foo = std::make_unique<Foo>();
// ... do some stuff, then return
}
In this example, we create a Foo object and assign it to our unique_ptr. Normally when you create something that's on the heap, you have to use new allocate space for it in the heap and construct it, and delete to destruct it and deallocate its space. Here, the unique_ptr handles deallocation as soon as you leave the scope where the unique_ptr was created (which in this case is the end of the function).
Only non-const unique_ptr can transfer the ownership of the managed object to another unique_ptr.
This means you can transfer the ownership of your pointer to a different unique_ptr, but only if it's not marked as const. Only one unique_ptr can own an object at a time.
One way to do that would be like this:
std::unique_ptr<Foo> foo1 = std::make_unique<Foo>();
std::unique_ptr<Foo> foo2 = std::move(foo1);
Now the pointer in foo1 has been moved to foo2. foo1 is no longer managing that memory, foo2 is.
The lifetime of an object managed by const std::unique_ptr is limited to the scope in which the pointer was created.
This means that when your unique_ptr leaves scope, it deletes the object that it points to. It's as if you did this:
void MyFunc()
{
Foo* foo = new Foo()
// ... do some stuff, then return
delete foo;
}
The benefit is that now you don't have to manually call delete, which is good because that's a potential memory leak if you forget to delete it.
Related
We have a function that returns a new allocated object as a output argument (ref to pointer).
MyFunc(MyObject*& obj)
{
obj = new MyObject();
}
Which is called like so:
Object* obj;
MyFunc(obj);
Internally the function does quite a bit and uses shared_ptr for memory management. When it is done, the object we would like to return is referenced by a shared_ptr. I am struggling on how to return our new allocated object as a regular pointer.
We would like to continue to use shared_ptr internally to reduce risks, but it does not seem to make sense to return a shared_ptr as the caller takes complete ownership over the returned value (the called function or object no longer needs or keeps a reference to the returned data) and we want them to have flexibility.
Does anyone have any suggestions for allowing us to use shared_ptr internally but have a regular pointer interface? Thanks
If for some reason you cannot/want not use std::unique_ptr or std::auto_ptr (for example if you need to have multiple owners internally during creation for some reason or your underlying methods require std::shared_ptr to be passed around), you can still make it work with std::shared_ptr by using custom deleter, as described here: https://stackoverflow.com/a/5995770/1274747
In the principle, after you're done before the return, you switch the deleter to not actually delete the instance (make the deleter "null") and then return by shared_ptr get(). Even after all shared_ptr objects are destroyed, the memory will not be deleted (as the nulled deleter will skip the deletion).
There is also a link in the comments not so well visible, which might be of your interest:
http://paste.ubuntu.com/23866812/
(not sure though if it would really work without the shared ownership of the switch in all cases, would need to test)
EDIT
As expected, with the linked simple disarmable deleter from the pastebin you need to be careful, because the deleter is actually copied for storing in std::shared_ptr.
But you can still make it work by using std::ref:
MyFunc(MyObject*& obj)
{
DisarmableDelete<MyObject> deleter;
std::shared_ptr<MyObject> ptr(new MyObject(), std::ref(deleter));
// do what is necessary to setup the object - protected by deleter
// ...
// disarm before return
deleter._armed = false;
obj = ptr.get();
// deleter disarmed - object not freed
}
And just for completeness (and to avoid a potential future broken link), here is the implementation of DisarmableDelete from http://paste.ubuntu.com/23866812/.
template <typename T, typename Deleter = typename std::default_delete<T> >
struct DisarmableDelete : private Deleter {
void operator()(T* ptr) { if(_armed) Deleter::operator()(ptr); }
bool _armed = true;
};
It depends on who "owns" the pointer, once it has been exposed to the 'outside world.' Ownership essentially boils down to: "who is responsible for freeing this memory, later?"
It can be answered with a simple question: when MyFunc is called, is the caller responsible for deleting the pointer when it's done?
If so, then MyFunc needs to 'release' the ownership, otherwise the shared_ptr will automatically delete the pointer, when it goes out of scope. This actually can't be done, using shared_ptr. You need to use a unique_ptr instead, and call unique_ptr::release().
If not - if MyFunc will simply use the resulting pointer and forget about it without delete-ing it - then you can simply return the 'raw' pointer using shared_ptr::get(). You must be careful, because this implies that the shared_ptr still exists elsewhere in your code.
I can see four alternatives, as highlighted below. They are all horrible, and short of switching your ownership to std::unique_ptr<T> and returning via obj = ptr.release(); I can only offer a hack where the argument is assigned to the pointer upon destruction, but you still need to catch the exception and test whether the pointer was assigned.
#include <iostream>
#include <memory>
#include <exception>
struct foo {
void bar() const { std::cout << this << " foo::bar()\n"; }
~foo() { std::cout << this << " deleted\n"; }
};
void f1(foo*& obj) {
obj = new foo;
// do stuff... if an exception is thrown before we return we are
// left with a memory leak
}
void f2(foo*& obj) {
auto holder = std::make_shared<foo>();
// do stuff.. if an exception is thrown the pointer will be
// correclty deleted.
obj = holder.get(); // awesome, I have a raw pointer!
} // oops, the destructor gets called because holder went out of
// scope... my pointer points to a deleted object.
void f3(foo*& obj) {
auto holder = std::make_unique<foo>();
// do stuff.. if an exception is thrown the pointer will be
// correclty deleted.
obj = holder.release(); // awesome, I have a raw pointer!
} // no problem whem holder goes out of scope because it does not own the pointer
void f4(foo*& obj) {
// a super-weird hack that assigns obj upon deletion
std::shared_ptr<foo> holder(new foo, [&obj](foo*& p){ obj = p; });
throw std::exception();
} // no problem whem holder goes out of scope because it does not own
// the pointer... but if an execption is throw we need to delete obj
int main() {
foo* p1;
f1(p1);
p1->bar();
foo* p2;
f2(p2);
// p2->bar(); // error
foo* p3;
f3(p3);
p3->bar();
foo* p4;
try {
f4(p4);
} catch(...) {
std::cout << "caught an exception... test whether p4 was assigned it\n";
}
p4->bar(); // I still need to delete this thing
}
The point of shared_ptr is to express shared ownership.
Anything that does not share in the ownership of an object -- that doesn't have the right to make an object lifetime last longer, and the object lifetime is the union of the shared owners request for object lifetime -- should not use a shared_ptr.
Here, you have a shared_ptr and you are going to return it. At that point, you are violating the assumptions of shared_ptr; anyone who kept a shared_ptr copy expects that its content can last as long as it requests.
Meanwhile, the calling code thinks it owns the MyObject* raw pointer you passed it.
This is an example of misuse of shared_ptr.
Saying "we have memory management issues, use shared_ptr" doesn't fix memory management issues. Proper use of shared_ptr requires care and design, and the design must be that when the end of the lifetime of the object in question is shared by 2 or more pieces of data and/or code.
The internal code, if it does not own the pointer, should use either something like an observer_ptr<T> or a raw T* (the first to make it clear it doesn't own the object).
Ownership should be explicit, and in a unique_ptr. It can then call .release() to pass ownership to a raw pointer if required; in practice, I would change your signature to take a unique_ptr&, or have it return a unique_ptr.
The caller would then call .release() when they want to use some other object lifetime management system, or that object lifetime management system should consume unique_ptrs (thus being extremely clear about taking ownership of things).
Use a non-hack solution.
Such as a std::shared_ptr<std::unique_ptr<T>>. In this case, you have shared ownership of a unique ownership.
The unique_ptr can have its ownership taken from it (via .release()). When it does so, all of the shared_ptrs that still exist will have their unique_ptr also be cleared.
This places the shared unique ownership front and center, instead of hacking a non-deleter into a shared_ptr and having dangling shared_ptrs that think they have ownership over data but do not.
The best approach is to use internally unique_ptr and call its .release() method before returning the raw pointer.
If you are stuck to use shared_ptr internally, an option is to create it specifying a custom, "noop" deleter, that just does nothing when shared_ptr is destroyed (instead of calling delete on the owned pointer). The get the raw pointer from shared_ptr as usual (.get() method).
An example of such a deleter can be found in the Boost library (null_deleter).
But please note that doing this effectively "disable" the usefulness of having a shared_ptr at all...
If your managing the lifespan of the allocated object internally with std::shared_ptr, and are returning a raw pointer for access and don't want this pointer to affect the ref count, you can return the raw pointer by calling shared_ptr.get().
It can be problematic to return smart pointers if your using a tool like Swig to generate wrappers for other languages.
Could someone explain simply the reason why this does not work:
std::shared_pointer<Bar> getSharedPointer() {
return std::make_shared<Bar>();
}
...
auto foo = getSharedPointer().get();
Apparently using the raw pointer foo will cause a segfault because the lifetime of the shared pointer returned by getSharedPointer() will have run out. Somehow I would expect it to last until the end of its scope (like whatever block it is inside).
Is this correct and are there any analogous examples to this situation?
For getSharedPointer().get();, getSharedPointer() returns a temporary std::shared_ptr which will be destroyed after the expression immediately, and the pointer managed by it will be deleted too. After that foo will become dangled, any dereference on it causes UB.
auto foo = getSharedPointer().get();
// foo have become dangled from here
You could use a named variable instead:
auto spb = getSharedPointer();
auto foo = spb.get();
// It's fine to use foo now, but still need to note its lifetime
// because spb will be destroyed when get out of its scope
// and the pointer being managed will be deleted too
auto foo = getSharedPointer().get();
Whenever a function returns a type that is not a reference, the result of calling the function is an rvalue. Also, because the function getSharedPointer() returns a class type, the result is a temporary object.
The lifetime of that temporary object is defined as the end of the evaluation of the outermost expression, here getSharedPointer().get(). As soon as the foo variable is initialized, the owning smart pointer is destroyed; when the last shared_ptr owning that object is destroyed, the object is deleted.
Here getSharedPointer() always returns shared_ptr that doesn't share the managed object (use_count() is 1), so when that copy of the last shared_ptr is destroyed, the object is destroyed and the pointer to the object is invalid.
(I am not sure why you are returning a shared_ptr and not a unique_ptr here.)
The proper use of a smart pointer, or any class that "owns" (controls the lifetime of) other resources (resources that you are still allowed to access directly), is to keep the "smart" pointer/owner alive as long as you need to access the ressource.
So the "smart" pointer (owning object) needs to be named. Also, I am not sure that you really would want to hide the fact that it is a smart pointer from the view of the reader with auto.
std::shared_pointer<Bar> foo = getSharedPointer();
// use foo.get()
You may want to hide the exact type of managed object:
std::shared_pointer<auto> foo = getSharedPointer();
Is that an accurate description of it? Does it make sense?
Are you guaranteed that the object it points to won't be deleted until the unique_ptr goes out of scope [even if you're not using the unique_ptr]?
Yes, std::unique_ptr follows the RAII design principle.
No, std::unique_ptr does not prevent other code from doing something stupid, like calling delete on a pointer that belongs to the unique_ptr. The unique_ptr itself will call a deleter1 on the object it owns when either:
it goes out of scope
or
the unique_ptr is reassigned (via operator= or reset) to point to a different object
One can also revoke unique_ptr's ownership of an object by moving to a different smart pointer, or using the release member function. This breaks the association between the object and the unique_ptr and unique_ptr will no longer clean up the object.
1 The default deleter will use either delete or delete [], depending on whether the target has array type. But unique_ptr is a template and its deleter can be customized, for example the cleanup operation for a FILE* can be chosen to be a call to fclose.
This capability can be used to schedule an arbitrary cleanup action to take place when the unique_ptr goes out of scope. RAII is used for keeping locks, closing files, and so forth -- clearly there would be major problems if the cleanup action were performed early just because the compiler didn't see any future usage of the smart pointer. Luckily the C++ object lifetime rules are completely deterministic (even the order of destruction for multiple automatic variables in the same scope is well defined), and you can count on the smart pointer cleaning up its owned object exactly when the smart pointer is itself destroyed.
std::unique_ptr is RAII, as the creation of the object also initializes it.
Are you guaranteed that the object it points to won't be deleted until the unique_ptr goes out of scope [even if you're not using the unique_ptr]?
If you make a few assumptions:
You don't move the unique_ptr (that would cause the newly moved to location to be the one that will delete
You don't delete the pointer yourself or add it to another control structure.
Or put more succinctly, yes it will live that long.
It is an application of RAII, indeed.
But you are not granted that the object it points to is not deleted by someone else before the unique_ptr goes out of scope.
E.g.
int* p = new int;
int* cp = p;
std::unique_ptr<int> up(p);
delete cp;
will give undefined behavior.
std::unique_ptr can be used for RAII, but doesn't prevent you from doing something like this:
#include <memory>
class Foo{};
int main()
{
Foo* pFoo = new Foo;
std::unique_ptr<Foo> upFoo(pFoo);
delete pFoo; // WRONG, double deletion when upFoo's destructor is called
}
Usually, the best way to use smart pointers is to pass the raw pointer as their constructor or make-like function, such as
#include <memory>
class Foo{};
int main()
{
std::unique_ptr<Foo> upFoo(new Foo); // or make_unique in C++14
}
I try to understand how std::unique_ptr works and for that I found this document. The author starts from the following example:
#include <utility> //declarations of unique_ptr
using std::unique_ptr;
// default construction
unique_ptr<int> up; //creates an empty object
// initialize with an argument
unique_ptr<int> uptr (new int(3));
double *pd= new double;
unique_ptr<double> uptr2 (pd);
// overloaded * and ->
*uptr2 = 23.5;
unique_ptr<std::string> ups (new std::string("hello"));
int len=ups->size();
What is confusing to me is that in this line
unique_ptr<int> uptr (new int(3));
We use integer as an argument (between round brackets) and here
unique_ptr<double> uptr2 (pd);
we used an pointer as an argument. Does it make any difference?
What is also not clear to me, is how pointers, declared in this way will be different from the pointers declared in a "normal" way.
The constructor of unique_ptr<T> accepts a raw pointer to an object of type T (so, it accepts a T*).
In the first example:
unique_ptr<int> uptr (new int(3));
The pointer is the result of a new expression, while in the second example:
unique_ptr<double> uptr2 (pd);
The pointer is stored in the pd variable.
Conceptually, nothing changes (you are constructing a unique_ptr from a raw pointer), but the second approach is potentially more dangerous, since it would allow you, for instance, to do:
unique_ptr<double> uptr2 (pd);
// ...
unique_ptr<double> uptr3 (pd);
Thus having two unique pointers that effectively encapsulate the same object (thus violating the semantics of a unique pointer).
This is why the first form for creating a unique pointer is better, when possible. Notice, that in C++14 we will be able to do:
unique_ptr<int> p = make_unique<int>(42);
Which is both clearer and safer. Now concerning this doubt of yours:
What is also not clear to me, is how pointers, declared in this way will be different from the pointers declared in a "normal" way.
Smart pointers are supposed to model object ownership, and automatically take care of destroying the pointed object when the last (smart, owning) pointer to that object falls out of scope.
This way you do not have to remember doing delete on objects allocated dynamically - the destructor of the smart pointer will do that for you - nor to worry about whether you won't dereference a (dangling) pointer to an object that has been destroyed already:
{
unique_ptr<int> p = make_unique<int>(42);
// Going out of scope...
}
// I did not leak my integer here! The destructor of unique_ptr called delete
Now unique_ptr is a smart pointer that models unique ownership, meaning that at any time in your program there shall be only one (owning) pointer to the pointed object - that's why unique_ptr is non-copyable.
As long as you use smart pointers in a way that does not break the implicit contract they require you to comply with, you will have the guarantee that no memory will be leaked, and the proper ownership policy for your object will be enforced. Raw pointers do not give you this guarantee.
There is no difference in working in both the concepts of assignment to unique_ptr.
int* intPtr = new int(3);
unique_ptr<int> uptr (intPtr);
is similar to
unique_ptr<int> uptr (new int(3));
Here unique_ptr automatically deletes the space occupied by uptr.
how pointers, declared in this way will be different from the pointers declared in a "normal" way.
If you create an integer in heap space (using new keyword or malloc), then you will have to clear that memory on your own (using delete or free respectively).
In the below code,
int* heapInt = new int(5);//initialize int in heap memory
.
.//use heapInt
.
delete heapInt;
Here, you will have to delete heapInt, when it is done using. If it is not deleted, then memory leakage occurs.
In order to avoid such memory leaks unique_ptr is used, where unique_ptr automatically deletes the space occupied by heapInt when it goes out of scope. So, you need not do delete or free for unique_ptr.
Unique pointers are guaranteed to destroy the object they manage when they go out of scope.
http://en.cppreference.com/w/cpp/memory/unique_ptr
In this case:
unique_ptr<double> uptr2 (pd);
pd will be destroyed when uptr2 goes out of scope. This facilitates memory management by automatic deletion.
The case of unique_ptr<int> uptr (new int(3)); is not different, except that the raw pointer is not assigned to any variable here.
From cppreference, one of the std::unique_ptr constructors is
explicit unique_ptr( pointer p ) noexcept;
So to create a new std::unique_ptr is to pass a pointer to its constructor.
unique_ptr<int> uptr (new int(3));
Or it is the same as
int *int_ptr = new int(3);
std::unique_ptr<int> uptr (int_ptr);
The different is you don't have to clean up after using it. If you don't use std::unique_ptr (smart pointer), you will have to delete it like this
delete int_ptr;
when you no longer need it or it will cause a memory leak.
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.