I have a class with a unique_ptr member.
class Foo {
private:
std::unique_ptr<Bar> bar;
...
};
The Bar is a third party class that has a create() function and a destroy() function.
If I wanted to use a std::unique_ptr
with it in a stand alone function I could do:
void foo() {
std::unique_ptr<Bar, void(*)(Bar*)> bar(create(), [](Bar* b){ destroy(b); });
...
}
Is there a way to do this with std::unique_ptr
as a member of a class?
This question is related to
c++
c++11
move-semantics
unique-ptr
You can simply use std::bind
with a your destroy function.
std::unique_ptr<Bar, std::function<void(Bar*)>> bar(create(), std::bind(&destroy,
std::placeholders::_1));
But of course you can also use a lambda.
std::unique_ptr<Bar, std::function<void(Bar*)>> ptr(create(), [](Bar* b){ destroy(b);});
You just need to create a deleter class:
struct BarDeleter {
void operator()(Bar* b) { destroy(b); }
};
and provide it as the template argument of unique_ptr
. You'll still have to initialize the unique_ptr in your constructors:
class Foo {
public:
Foo() : bar(create()), ... { ... }
private:
std::unique_ptr<Bar, BarDeleter> bar;
...
};
As far as I know, all the popular c++ libraries implement this correctly; since BarDeleter
doesn't actually have any state, it does not need to occupy any space in the unique_ptr
.
You know, using a custom deleter isn't the best way to go, as you will have to mention it all over your code.
Instead, as you are allowed to add specializations to namespace-level classes in ::std
as long as custom types are involved and you respect the semantics, do that:
Specialize std::default_delete
:
template <>
struct ::std::default_delete<Bar> {
default_delete() = default;
template <class U>
constexpr default_delete(default_delete<U>) noexcept {}
void operator()(Bar* p) const noexcept { destroy(p); }
};
And maybe also do std::make_unique()
:
template <>
inline ::std::unique_ptr<Bar> ::std::make_unique<Bar>() {
auto p = create();
if (!p)
throw std::runtime_error("Could not `create()` a new `Bar`.");
return { p };
}
Unless you need to be able to change the deleter at runtime, I would strongly recommend using a custom deleter type. For example, if use a function pointer for your deleter, sizeof(unique_ptr<T, fptr>) == 2 * sizeof(T*)
. In other words, half of the bytes of the unique_ptr
object are wasted.
Writing a custom deleter to wrap every function is a bother, though. Thankfully, we can write a type templated on the function:
Since C++17:
template <auto fn>
using deleter_from_fn = std::integral_constant<decltype(fn), fn>;
template <typename T, auto fn>
using my_unique_ptr = std::unique_ptr<T, deleter_from_fn<fn>>;
// usage:
my_unique_ptr<Bar, destroy> p{create()};
Prior to C++17:
template <typename D, D fn>
using deleter_from_fn = std::integral_constant<D, fn>;
template <typename T, typename D, D fn>
using my_unique_ptr = std::unique_ptr<T, deleter_from_fn<D, fn>>;
// usage:
my_unique_ptr<Bar, decltype(destroy), destroy> p{create()};
It's possible to do this cleanly using a lambda in C++11 (tested in G++ 4.8.2).
Given this reusable typedef
:
template<typename T>
using deleted_unique_ptr = std::unique_ptr<T,std::function<void(T*)>>;
You can write:
deleted_unique_ptr<Foo> foo(new Foo(), [](Foo* f) { customdeleter(f); });
For example, with a FILE*
:
deleted_unique_ptr<FILE> file(
fopen("file.txt", "r"),
[](FILE* f) { fclose(f); });
With this you get the benefits of exception-safe cleanup using RAII, without needing try/catch noise.
Source: Stackoverflow.com