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How do I use a custom deleter with a std::unique_ptr member?

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Can you delete a unique pointer?

An explicit delete for a unique_ptr would be reset() . But do remember that unique_ptr are there so that you don't have to manage directly the memory they hold. That is, you should know that a unique_ptr will safely delete its underlying raw pointer once it goes out of scope.

What is a Unique_ptr?

std::unique_ptr is a smart pointer that owns and manages another object through a pointer and disposes of that object when the unique_ptr goes out of scope. The object is disposed of, using the associated deleter when either of the following happens: the managing unique_ptr object is destroyed.


Assuming that create and destroy are free functions (which seems to be the case from the OP's code snippet) with the following signatures:

Bar* create();
void destroy(Bar*);

You can write your class Foo like this

class Foo {

    std::unique_ptr<Bar, void(*)(Bar*)> ptr_;

    // ...

public:

    Foo() : ptr_(create(), destroy) { /* ... */ }

    // ...
};

Notice that you don't need to write any lambda or custom deleter here because destroy is already a deleter.


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.


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.


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()};

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 };
}