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Let's say I want to use a custom deleter with an unique_ptr:
void custom_deleter(int* obj)
{
delete obj;
}
Why do I have to write this:
std::unique_ptr<int, void(*)(int*)> x(new int, custom_deleter);
instead of this:
std::unique_ptr<int> x(new int, custom_deleter); //does not compile
?
Can't the type of the deleter be inferred?
699
An unique_ptr has exclusive ownership of the object it points to and will destroy the object when the pointer goes out of scope.
When to use unique_ptr? Use unique_ptr when you want to have single ownership(Exclusive) of the resource. Only one unique_ptr can point to one resource. Since there can be one unique_ptr for single resource its not possible to copy one unique_ptr to another.
yes, it's an RAII class.
A unique_ptr can only be moved. This means that the ownership of the memory resource is transferred to another unique_ptr and the original unique_ptr no longer owns it. We recommend that you restrict an object to one owner, because multiple ownership adds complexity to the program logic.
For unique_ptr, the deleter is part of the type:
template <
class T,
class Deleter = std::default_delete<T>
> class unique_ptr;
As such, when you're constructing an object, you need to specify its type. The line you're writing:
std::unique_ptr<int> x(new int, custom_deleter);
is equivalent to:
std::unique_ptr<int, std::default_delete<int> > x(new int, custom_deleter);
And you cannot construct a std::default_delete<int> from custom_deleter.
The only way to infer the deleter type is to use template deduction on that part too:
template <typename T, typename Deleter>
std::unique_ptr<T, Deleter> make_unique_ptr(T* ptr, Deleter deleter) {
return std::unique_ptr<T, Deleter>(ptr, deleter);
}
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