问题
I want to keep the smart behavior of std::shared_ptr
. So is there a way to cast a shared void pointer to another type while without confusing the reference counting? I can't get the raw pointer and create a new shared pointer from it.
回答1:
You can use the pointer casts from rob mayoff's answer; but be careful. It is easy to unintentionally trigger undefined behavior here:
struct MyClass {};
void* rawPtr = new MyClass;
shared_ptr<void> exampleVoid(rawPtr); // Undefined behavior;
// calls delete (void*)ptr;
shared_ptr<void> exampleVoidCons(new MyClass);
// OK, calls shared_ptr<void>::shared_ptr<MyClass>(MyClass*) which
// makes a deleter calling delete (MyClass*)ptr;
shared_ptr<MyClass> example(new MyClass); // OK, calls delete (MyClass*)ptr;
shared_ptr<void> castToVoid = static_pointer_cast<void>(example);
// OK, shared_ptr's deleter is erased so this still calls delete (MyClass*)ptr;
Typically this undefined behavior will result in the type's destructor not being called. For example, see the output on ideone and note that the version put into a void*
never prints that it was destroyed.
See C++11 5.3.5 [expr.delete]/3:
In the first alternative (delete object), if the static type of the object to be deleted is different from its dynamic type, the static type shall be a base class of the dynamic type of the object to be deleted and the static type shall have a virtual destructor or the behavior is undefined.
Since the actual object will never have a dynamic type void
, and void
is never a base class of a dynamic type, delete
ing a void*
triggers undefined behavior.
回答2:
You can use std::static_pointer_cast or std::dynamic_pointer_cast depending on what kind of cast you want.
来源:https://stackoverflow.com/questions/25858939/is-there-a-way-to-cast-shared-ptrvoid-to-shared-ptrt