Problem:
Let's assume I have an algorithm that takes a unique_ptr to some type:
void FancyAlgo(unique_ptr<SomeType>& ptr);
Now I have shared_ptr sPtr to SomeType, and I need to apply the same algorithm on sPtr. Does this mean I have to duplicate the algorithm just for the shared_ptr?
void FancyAlgo(shared_ptr<SomeType>& sPtr);
I know smart pointers come with ownership of the underlying managed object on the heap. Here in my FancyAlgo, ownership is usually not an issue. I thought about stripping off the smart pointer layer and do something like:
void FancyAlgo(SomeType& value);
and when I need to call it with unique_ptr:
FancyAlgo(*ptr);
likewise for shared_ptr.
1, Is this an acceptable style in PRODUCTION code?(I saw somewhere that in a context of smart pointers, you should NOT manipulate raw pointers in a similar way. It has the danger of introducing mysterious bugs.)
2, Can you suggest any better way (without code duplication) if 1 is not a good idea.
Thanks.
Smart pointers are about ownership. Asking for a smart pointer is asking for ownership information or control.
Asking for a non-const lvalue reference to a smart pointer is asking for permission to change the ownership status of that value.
Asking for a const lvalue reference to a smart pointer is asking for permission to query the ownership status of that value.
Asking for an rvalue reference to a smart pointer is being a "sink", and promising to take that ownership away from the caller.
Asking for a const rvalue reference is a bad idea.
If you are accessing the pointed to value, and you want it to be non-nullable, a reference to the underlying type is good.
If you want it to be nullable, a boost::optional<T&> or a T* are acceptable, as is the std::experimental "world's dumbest smart pointer" (or an equivalent hand-written one). All of these are non-owning nullable references to some variable.
In an interface, don't ask for things you don't need and won't need in the future. That makes reasoning about what the function does harder, and leads to problems like you have in the OP. A function that reseats a reference is a very different function from one that reads a value.
Now, a more interesting question based off yours is one where you want the function to reseat the smart pointer, but you want to be able to do it to both shared and unique pointer inputs. This is sort of a strange case, but I could imagine writing a type-erase-down-to-emplace type (a emplace_sink<T>).
template<class T>
using later_ctor = std::function<T*(void*)>;
template<class T, class...Args>
later_ctor<T> delayed_emplace(Args&&...args) {
// relies on C++1z lambda reference reference binding, write manually
// if that doesn't get in, or don't want to rely on it:
return [&](void* ptr)->T* {
return new T(ptr)(std::forward<Args>(args));
};
}
namespace details {
template<class T>
struct emplace_target {
virtual ~emplace_target() {}
virtual T* emplace( later_ctor<T> ctor ) = 0;
};
}
template<class T>
struct emplacer {
std::unique_ptr<emplace_target<T>> pImpl;
template<class...Args>
T* emplace( Args&&... args ) {
return pImpl->emplace( delayed_emplace<T>(std::forward<Args>(args)...) );
}
template<class D>
emplacer( std::shared_ptr<T, D>& target ):
pImpl( new details::emplace_shared_ptr<T,D>(&target) ) // TODO
{}
template<class D>
emplacer( std::unique_ptr<T, D>& target ):
pImpl( new details::emplace_unique_ptr<T,D>(&target) ) // TODO
{}
};
etc. Lots of polish needed. The idea is to type-erase construction of an object T into an arbitrary context. We might need to special case shared_ptr so we can call make_shared, as a void*->T* delayed ctor is not good enough to pull that off (not fundamentally, but because of lack of API hooks).
Aha! I can do a make shared shared ptr without special casing it much.
We allocate a block of memory (char[sizeof(T)]) with a destructor that converts the buffer to T then calls delete, in-place construct in that buffer (getting the T*), then convert to a shared_ptr<T> via the shared_ptr<T>( shared_ptr<char[sizeof(T)]>, T* ) constructor. With careful exception catching this should be safe, and we can emplace using our emplacement function into a make_shared combined buffer.
Related
I am implementing a command pattern implementations with large number of actions and parameters involved. To simplify I am planning to use class that can hold all possible params to module in a map.
class ParamBag {
public:
add(int paramId, shared_ptr<IParam> param);
bool contains(int paramId);
std::shared_ptr<IParam> get(paramId);
private:
int mask;
std::map<int, std::shared_ptr<IParam>> params;
};
One clear downside of this implementation is each param has to extend from IParam interface, can I somehow simplify this.
If the one that uses the param after the get knows the type of the param, then you can use c++17 std::any, or if you must use c++11 you can try boost::any, or if none of those you can resort back to a void*.
The difference is that void* will not fail on a cast to a wrong type, where any_cast would throw an exception, or return nullptr if used with a pointer. You would also need to use a custom deleter in the std::shared_ptr in order to be able to free the void*.
A library requires binary data to be shared as void *.
The data to be shared is available as shared_ptr<T>.
Is there a way to cast shared_ptr<T> to void * ?
PS: Static casting does not work:
error: invalid static_cast from type ‘std::shared_ptr<DataPacket>’ to type ‘void*’
static_cast<void *>(binData);
You're going about this wrong. Your problem is not "I need to interpret a shared_ptr<T> as a void*" — your problem is "I have a smart pointer to an object, and I need a dumb pointer to that object".
shared_ptr<T> has a function that does exactly that.
shared_ptr<T> smart;
// ... some code here points smart at an object ...
T *dumb1 = smart.get(); // creates a dumb pointer to the object managed by smart
void *dumb2 = smart.get(); // dumb pointers automatically convert to void*
Do be careful to note that the dumb pointer this creates does not participate in the shared ownership scheme, so you have to take care to ensure the lifetime of the object lasts as long as you need it to.
I can't understand the function move in c++11.
From here, I got things below:
Although note that -in the standard library- moving implies that the
moved-from object is left in a valid but unspecified state. Which
means that, after such an operation, the value of the moved-from
object should only be destroyed or assigned a new value; accessing it
otherwise yields an unspecified value.
In my opinion, after move(), the moved-from object has been "clear". However, I've done a test below:
std::string str = "abcd";
std::move(str);
std::cout<<str;
I got abcd on my screen.
So has the str been destroyed? If so, I could get abcd because I'm just lucky? Or I misunderstood the function move?
Besides, when I read C++ Primer, I got such a code:
class Base{/* ... */};
class D: public Base{
public:
D(D&& d): Base(std::move(d)){/* use d to initialize the members of D */}
};
I'm confused now. If the function move will clear the object, the parameter d will be clear, how could we "use d to initialize the members of D"?
std::move doesn't actually do anything. It's roughly analogous to a cast expression, in that the return value is the original object, but treated differently.
More precisely, std::move returns the object in a form which is amenable to its resources being 'stolen' for some other purpose. The original object remains valid, more or less (you're only supposed to do certain special things to it, though that's primarily a matter of convention and not necessarily applicable to non-standard-library objects), but the stolen-away resources no longer belong to it, and generally won't be referenced by it any more.
But! std::move doesn't, itself, do the stealing. It just sets things up for stealing to be allowed. Since you're not doing anything with the result, let alone something which could take advantage of the opportunity, nothing gets stolen.
std::move doesn’t move anything. std::move is merely a function template that perform casts. std::move unconditionally casts its argument to an rvalue,
std::move(str);
With this expression you are just doing type cast from lvalue to rvalue.
small modification in program to understand better.
std::string str = "abcd";
std::string str1 = std::move(str);
std::cout<<str<<std::endl;
std::cout<<str1<<std::endl;
str lvalue typecast to rvalue by std::move, std::string = std::move(str); =>this expression call the string move constructor where actual stealing of resources take placed. str resources(abcd) are steeled and printed empty string.
Here is sample implementation of move function. Please note that it is not complete implementation of standard library.
template<typename T> // C++14; still in
decltype(auto) move(T&& param) // namespace std
{
using ReturnType = remove_reference_t<T>&&;
return static_cast<ReturnType>(param);
}
Applying std::move to an object tells the compiler that the object is eligible to be moved from. It cast to the rvalue.
class Base{/* ... */};
class D: public Base{
public:
D(D&& d): Base(std::move(d)){/* use d to initialize the members of D */}
};
Base(std::move(d)) it will do up-casting only move the base class part only.
Here one more interesting thing to learn for you. If you do not invoke base class destructor with std::move like D(D&& d): Base(d) then d will be considered as lvalue and copy constructor of Base class involved instead of move constructor. Refer for more detail Move constructor on derived object
I have the following member variable in a class:
std::vector<std::unique_ptr<Object>> objects_;
I explicitly want the vector to maintain ownership at all times. I've seen suggestions that in order for a member function to access a pointer in the vector and make changes to the object T wrapped in the std::unique_ptr, we must move the pointer to calling code, i.e:
void foo(int i) {
auto object = std::move( vector.at( i ) ); // move object to caller (caller owns)
object->dosomething();
vector.at(i) = std::move(object); // move back into vector (vector owns)
}
Another method was to work with raw pointers:
void foo(int i) {
Object* object = vector.at( i ).get();
object->doSomething();
}
However, I've been working with this:
void foo(int i) {
auto& object = vector.at( i );
object->doSomething();
}
Which is the correct and most robust method for my case? Does this function ever take ownership of the data in the std::unique_ptr? Is there a way to access Object without playing with the std::unique_ptr?
(excuse me if my methods are incorrect, I hope I got the point across)
The first approach will not retain ownership of the object if object->dosomething() throws an exception (i.e. it is not exception safe) since the second std::move() statement will not be executed.
Assuming C++11, both of the other approaches are effectively equivalent, subject to the assumption that the owned pointer is not null. Under the same assumption, the code can be simplified to
void foo(int i)
{
vector.at(i)->doSomething();
}
which will work with all C++ standards (not just C++11 or later).
It is possible to access the object without monkeying with the unique_ptr - simply store the pointer elsewhere and use that. However, that does compromise the purpose of using std::unique_ptr in the first place. And is error-prone - for example, the std::unique_ptr can destroy the object, and leave those other pointers dangling.
If you are really that worried about the potential of your vector losing ownership, consider using a shared_ptr instead.
Consider a class D derived from a class B, and a sb instance of std::shared_ptr<B>. Once I have verified that dynamic_cast<D*>(sb.get()) is possible, I want to create a proper std::shared_ptr<D> from sb. In other words, I'd like to implement a kind of dynami_cast<> between shared_ptr's. How can I do this in a clean way? A possible solution would be to make B derive from std::enable_shared_from_this and to use shared_from_this() from the (casted) pointer to D. But this would require a change in the definition of type B. Are there better ideas? Is there anything in boost?
Are you aware of std::dynamic_pointer_cast?
http://en.cppreference.com/w/cpp/memory/shared_ptr/pointer_cast
As other answers pointed out, the standard library already provides what you want, but for completeness, implementing it is easy:
template<typename To, typename From>
std::shared_ptr<To>
dynamic_pointer_cast(const std::shared_ptr<From>& from)
{
return std::shared_ptr<To>(from, dynamic_cast<To*>(from.get()));
}
This uses the aliasing constructor to create a new shared_ptr of a different type that shares ownership with from, so they share the same reference count even though they own different pointers (in this case the pointers they own are different types but point to the same object, but that doesn't have to be true.)
This isn't quite right, because if the cast fails (returning a null pointer) you get a shared_ptr that stores a null pointer, but shares ownership with a non-null pointer. To handle that we need a small tweak:
template<typename To, typename From>
std::shared_ptr<To>
dynamic_pointer_cast(const std::shared_ptr<From>& from)
{
if (auto p = dynamic_cast<To*>(from.get()))
return std::shared_ptr<To>(from, p);
return {};
}
Now we return an empty shared_ptr<To> if the cast fails, which is a better match to the behaviour of dynamic_cast<To*>.
Look here: dynamic_pointer_cast