c++ - boost const_iterator 到迭代器的转换

Question

我尝试了两种方法来实现从 const_iterator 到迭代器的转换。所有迭代器都基于 boost/iterator .

方法 1 定义了一个 iterator<T>类(class)。 iterator<const T>将代表 const_iterator . iterator<T>有一个返回 iterator<const T> 的转换运算符.这对模板函数失败，因为在模板实例化期间不会发生类型转换。

方法 2 在理论上可行。实际上，我需要为 iterator<T> 定义每个方法。 :

#include <iostream>
#include <boost/iterator/iterator_adaptor.hpp>
#include <vector>

template<typename Container>
class Cit
        : public boost::iterator_adaptor<
                Cit<Container>, // Derived
                typename Container::const_iterator, // Base
                const typename Container::value_type> {
    using self_type = Cit<Container>;
    friend class boost::iterator_core_access;
public:
    explicit Cit(typename Container::const_iterator it)
            : self_type::iterator_adaptor_(it) {}
};


template<typename Container>
class It : public Cit<Container> {
protected:
    using reference = typename Container::reference;
    using self_type = It<Container>;
    using Base = Cit<Container>;
public:
    explicit It(typename Container::iterator it)
            : Base(it) {}

    reference operator*() const {
        return const_cast<reference>(Base::operator*());
    }
    // Try to hide every method from Cit<Container>
    // ... 
    // ... 
    // ... 
    // ... oh well.
private:
    friend class boost::iterator_core_access;
};

// A template function
template<typename Container>
void foo(Cit<Container> it_begin,
         Cit<Container> it_end) {
    for (auto it = it_begin; it != it_end; ++it) {
        std::cout << *it << "\n";
    }

}

int main() {
    typedef std::vector<int> Container;
    Container v = {0, 1, 2, 3};  // content array
    It<Container> it_begin(v.begin());
    It<Container> it_end(v.end());
    // Assert It can implicitly convert to Cit even during template 
    // instantiation.
    foo(it_begin, it_end);
    return 0;
}

这似乎否定了使用 boost/iterator 的好处.

Is there a better way to make iterator and const_iterator with boost/iterator?

方法一:

#include <iostream>
#include <boost/iterator/iterator_adaptor.hpp>
#include <vector>

template<typename Container>
class It
        : public boost::iterator_adaptor<
                It<Container>, // Derived
                typename Container::const_iterator, // Base
                typename std::conditional<std::is_const<Container>::value,
                        const typename Container::value_type,
                        typename Container::value_type
                        >::type // Value
        > {
    using self_type = It<Container>;
    friend class boost::iterator_core_access;
public:
    explicit It(typename Container::const_iterator it)
            : self_type::iterator_adaptor_(it) {}
};

template <typename C> using Cit = It<const C>;

// A template function
template<typename Container>
void foo(Cit<Container> it_begin,
         Cit<Container> it_end) {
    for (auto it = it_begin; it != it_end; ++it) {
        std::cout << *it << "\n";
    }

}

int main() {
    typedef std::vector<int> Container;
    Container v = {0, 1, 2, 3};  // content array
    It<Container> it_begin(v.begin());
    It<Container> it_end(v.end());
    // Assert It can implicitly convert to from Cit to It even
    // during template instantiation.
    foo(it_begin, it_end);
    return 0;
}

错误信息:

error: no matching function for call to ‘foo(It<std::vector<int> >&, It<std::vector<int> >&)’
     foo(it_begin, it_end);
                         ^
main.cpp:26:6: note: candidate: template<class Container> void foo(Cit<Container>, Cit<Container>)
 void foo(Cit<Container> it_begin,
      ^~~
main.cpp:26:6: note:   template argument deduction/substitution failed:
main.cpp:41:25: note:   types ‘const C’ and ‘std::vector<int>’ have incompatible cv-qualifiers
     foo(it_begin, it_end);

Answer 1

我会专门化模板:

template <typename T>
class MyIt : public boost::iterator_adaptor<MyIt<T>,              // Derived
                                            typename T::iterator, // Base
                                            typename T::reference> {
    friend class boost::iterator_core_access;

  public:
    static constexpr bool is_const = false;
    explicit MyIt(typename MyIt::base_type it) : MyIt::iterator_adaptor_(it) {}
};

template <typename T>
class MyIt<T const> : public boost::iterator_adaptor<MyIt<T const>,              // Derived
                                                     typename T::const_iterator, // Base
                                                     typename T::const_reference> {
    friend class boost::iterator_core_access;

  public:
    static constexpr bool is_const = true;

    explicit MyIt(typename MyIt::base_type it) : MyIt::iterator_adaptor_(it) {}
};

这里已经允许转换，但是如果你想有一个显式的“to-const-cast”很容易写:

template <typename T>
static MyIt<T const> make_const(MyIt<T> it) { return MyIt<T const>(it.base()); }

---

使用它:

// A template function
template <typename It> void foo(It it_begin, It it_end) {
    static_assert(It::is_const == std::is_const<typename std::remove_reference<decltype(*it_begin)>::type>::value, "mismatch");
    if (It::is_const)
        std::cout << "Const: ";

    for (auto it = it_begin; it != it_end; ++it)
        std::cout << *it << " ";

    std::cout << "\n";
}

如您所见，我们的函数并不关心特定的迭代器(这是迭代器的全部意义所在)。您可以将它与 const 和非常量一起使用:

template <typename C> void foo(C const &c) {
    MyIt<C const> b(c.begin()), e(c.end());
    foo(b, e);
}

template <typename C> void foo(C &c) {
    MyIt<C> b(c.begin()), e(c.end());
    foo(b, e);
}

Quick Demo Live On Coliru

 std::vector<int> v{ 0, 1, 2, 3 };
 foo(v);

 auto const &constv = v;
 foo(constv);

Prints

void foo(C&) [with C = std::vector<int>]
0 1 2 3 
void foo(const C&) [with C = std::vector<int>]
Const: 0 1 2 3

强制构造迭代器

这似乎是您的代码的内容。所以，让我们强制吧!这是 MyIt<C> 的简单更改至 MyIt<C const> :

template <typename C> void foo(C &c) {
    MyIt<C const> b(c.begin()), e(c.end()); //  <--- note the const
    foo(b, e);
}

现在 foo 将使用 const 迭代器调用，即使对于非 const C。如果你认为这很微妙，你可以使用上面显示的帮助器:

template <typename C> void foo(C &c) {
    MyIt<C> b(c.begin()), e(c.end()); 
    foo(make_const(b), make_const(e)); //  <--- now more explicit?
}

当然在foo您可以自由修改 static_assert所以它首先拒绝为非常量迭代器编译:

// A template function
template <typename It> void foo(It it_begin, It it_end) {
    static_assert(std::is_const<typename std::remove_reference<decltype(*it_begin)>::type>::value, "non-const disallowed");
    if (It::is_const)
        std::cout << "Const: ";

    for (auto it = it_begin; it != it_end; ++it)
        std::cout << *it << " ";

    std::cout << "\n";
}

您可以为任何 MyIt<> 添加重载做凝固:

template <typename C> 
typename std::enable_if<!std::is_const<C>::value>::type
foo(MyIt<C> b, MyIt<C> e) {
    foo(make_const(b), make_const(e));
}

所以，现在每次调用 foo被迫const模式。

完整 list

最后一个完整的演示:

Live On Coliru

#include <boost/iterator/iterator_adaptor.hpp>
#include <iostream>
#include <vector>

template <typename T>
class MyIt : public boost::iterator_adaptor<MyIt<T>,              // Derived
                                            typename T::iterator, // Base
                                            typename T::reference> {
    friend class boost::iterator_core_access;

  public:
    static constexpr bool is_const = false;
    explicit MyIt(typename MyIt::base_type it) : MyIt::iterator_adaptor_(it) {}
};

template <typename T>
class MyIt<T const> : public boost::iterator_adaptor<MyIt<T const>,              // Derived
                                                     typename T::const_iterator, // Base
                                                     typename T::const_reference> {
    friend class boost::iterator_core_access;

  public:
    static constexpr bool is_const = true;

    explicit MyIt(typename MyIt::base_type it) : MyIt::iterator_adaptor_(it) {}
};

template <typename T>
static MyIt<T const> make_const(MyIt<T> it) { return MyIt<T const>(it.base()); }

// A template function
template <typename It> void foo(It it_begin, It it_end) {
    static_assert(std::is_const<typename std::remove_reference<decltype(*it_begin)>::type>::value, "non-const disallowed");
    if (It::is_const)
        std::cout << "Const: ";

    for (auto it = it_begin; it != it_end; ++it)
        std::cout << *it << " ";

    std::cout << "\n";
}

template <typename C> 
typename std::enable_if<!std::is_const<C>::value>::type
foo(MyIt<C> b, MyIt<C> e) {
    foo(make_const(b), make_const(e));
}

template <typename C> void foo(C &c) {
    std::cout << __PRETTY_FUNCTION__ << "\n";
    MyIt<C> b(c.begin()), e(c.end());
    foo(b, e);
}

int main() {
    std::vector<int> v{ 0, 1, 2, 3 };
    foo(v);

    auto const &constv = v;
    foo(constv);
}

现在打印:

void foo(C&) [with C = std::vector<int>]
Const: 0 1 2 3 
void foo(C&) [with C = const std::vector<int>]
Const: 0 1 2 3