将STL容器<T*>转换为容器<T const*>

Question

我正在寻找一种方法来制定一个类:

• 一个接口,使用最大'constness'的指针STL容器
• 但是它在内部突变了指向的对象
• 与非const模拟相比,没有额外的运行时开销

理想情况下,与非const版本相比,解决方案将编译为无额外代码,因为const/non-const-ness只是对程序员的帮助.

这是我到目前为止所尝试的:

#include <list>
#include <algorithm>

using namespace std;
typedef int T;

class C
{
public:
    // Elements pointed to are mutable, list is not, 'this' is not - compiles OK
    list<T *> const & get_t_list() const { return t_list_; }

    // Neither elements nor list nor' this' are mutable - doesn't compile
    list<T const *> const & get_t_list2() const { return t_list_; }

    // Sanity check: T const * is the problem - doesn't compile
    list<T const *> & get_t_list3() { return t_list_; }

    // Elements pointed to are immutable, 'this' and this->t_list_ are
    // also immutable - Compiles OK, but actually burns some CPU cycles
    list<T const *> get_t_list4() const {
        return list<T const *>( t_list_.begin() , t_list_.end() );
    }

private:
    list<T *> t_list_;
};

如果没有类型转换的解决方案,我想了解如何制定具有所述属性的类的替代建议.

Answer 1

让我们假设您可以转换list<T*>&为list<T const *>&.现在考虑以下代码:

list<char*> a;
list<char const*>& b = a;

b.push_back("foo");

a.front()[0] = 'x'; // oops mutating const data

这与转换T**为的概念问题相同T const**.

如果要提供对基础数据的只读访问,您将需要提供一些它的自定义视图,可能使用自定义的迭代器.

像下面这样的东西.

template <typename It>
class const_const_iterator {
private:
    using underlying_value_type = typename std::iterator_traits<It>::value_type;

    static_assert(std::is_pointer<underlying_value_type>(),
                  "must be an iterator to a pointer");

    using pointerless_value_type = typename std::remove_pointer<underlying_value_type>::type;

public:
    const_const_iterator(It it) : it(it) {}

    using value_type = pointerless_value_type const*;

    value_type operator*() const {
        return *it; // *it is a T*, but we return a T const*,
                    // converted implicitly
                    // also note that it is not assignable
    }

    // rest of iterator implementation here
    // boost::iterator_facade may be of help

private:
    It it;
};

template <typename Container>
class const_const_view {
private:
    using container_iterator = typename Container::iterator;

public:
    using const_iterator = const_const_iterator<container_iterator>;
    using iterator = const_iterator;

    const_const_view(Container const& container) : container(&container) {}

    const_iterator begin() const { return iterator(container->begin()); }
    const_iterator end() const { return iterator(container->end()); }

private:
    Container const* container;
}