Question

相关：Why do standard containers require allocator_type::value_type to be the element type?

据说自C ++ 17以来已弃用以下内容：

template<>
struct allocator<void>;

我想知道它是否已被弃用，因为仅主模板现在能够容纳allocator<void>，或者不推荐使用allocator<void>的用例。

如果是后者，我想知道为什么。我认为allocator<void>在指定未绑定到特定类型的分配器时非常有用（所以只需要一些模式/元数据）。

Answer 1

根据p0174r0

同样，std::allocator<void>被定义为各种模板重新绑定技巧可以在原始的C ++ 98库中运行，但确实如此不是实际的分配器，因为它缺少allocate和deallocate 成员函数，默认情况下无法合成 allocator_traits。这种需求随着C ++ 11和void_pointer而消失了。和allocator_traits中的const_void_pointer类型别名。但是，我们继续指定它以避免破坏旧代码根据C ++ 11，尚未升级为支持通用分配器。

Answer 2

不是std::allocator<void>被弃用，不是明确的专长。

它过去看起来像：

template<class T>
struct allocator {
    typedef T value_type;
    typedef T* pointer;
    typedef const T* const_pointer;
    // These would be an error if T is void, as you can't have a void reference
    typedef T& reference;
    typedef const T& const_reference;

    template<class U>
    struct rebind {
        typedef allocator<U> other;
    }

    // Along with other stuff, like size_type, difference_type, allocate, deallocate, etc.
}

template<>
struct allocator<void> {
    typedef void value_type;
    typedef void* pointer;
    typedef const void* const_pointer;

    template<class U>
    struct rebind {
        typdef allocator<U> other;
    }
    // That's it. Nothing else.
    // No error for having a void&, since there is no void&.
}

现在，由于std::allocator<T>::reference和std::allocator<T>::const_reference已被弃用，因此void不需要明确的专业化。您仅可以使用std::allocator<void>以及std::allocator_traits<std::allocator<void>>::template rebind<U>来获得std::allocator<U>，就不能实例化std::allocator<void>::allocates。

例如：

template<class Alloc = std::allocator<void>>
class my_class;  // allowed

int main() {
    using void_allocator = std::allocator<void>;
    using void_allocator_traits = std::allocator_traits<void_allocator>;
    using char_allocator = void_allocator_traits::template rebind_alloc<char>;
    static_assert(std::is_same<char_allocator, std::allocator<char>>::value, "Always works");

    // This is allowed
    void_allocator alloc;

    // These are not. Taking the address of the function or calling it
    // implicitly instantiates it, which means that sizeof(void) has
    // to be evaluated, which is undefined.
    void* (void_allocator::* allocate_mfun)(std::size_t) = &void_allocator::allocate;
    void_allocator_traits::allocate(alloc, 1);  // calls:
    alloc.allocate(1);
}

std :: allocator <void>的弃用

2 个答案: