c++ - 使用未定义类型错误定义树

Question

我有一个 Visual Studio 2013 C++11 项目，我在其中定义了一个树状结构(其中迭代器的每个元素本身就是一个容器)。

struct some_tree;

class some_tree_iterator 
    : public boost::iterator_facade<
        some_tree_iterator,
        some_tree,
        std::forward_iterator_tag,
        some_tree>
{
public:
    // ...

private:
    // error C2027: use of undefined type 'some_tree'
    some_tree dereference() const { return some_tree( /*init values*/ ); };

    // ...
};

struct some_tree
{
    some_tree_iterator begin();
    some_tree_iterator end();
    std::string value() const;
};

int main(int argc, char* argv[])
{
    some_tree foo;
    for (auto x : foo)
    {
        std::cout << x.name() << std::endl;
        for (auto y : x)
        {
            std::cout << "\t" << y.value() << std::endl;
        }
    }

    return 0;
}

绕过使用未定义类型 问题的唯一方法是返回指向新some_tree 容器的指针吗？像这样:

std::shared_ptr<some_tree> some_tree_iterator::dereference() const 
{ 
    return std::make_shared<some_tree>( /* init values*/ ); 
};

Answer 1

两种方法:

方法一

在定义类之前，您不得引用该类。所以你必须这样做:

struct some_tree;

class some_tree_iterator 
{
public:
    // ...

private:
    some_tree dereference() const;

    // ...
};

struct some_tree
{
    some_tree_iterator begin();
    some_tree_iterator end();
    std::string value() const;
};

some_tree some_tree_iterator::dereference() const { return some_tree( /*init values*/ ); };

STL 方法:

此外，您也可以像 STL 一样在 struct some_tree 中定义迭代器(事实上，STL 的迭代器是模板，因此它们在类中定义，即使在其他地方声明也是如此):

struct some_tree
{
  class iterator 
    : public boost::iterator_facade<
        iterator,
        some_tree,
        std::forward_iterator_tag,
        some_tree>
  {
    public:

    private:
      some_tree dereference() const { return some_tree( /*init values*/ ); };

  };

  iterator begin();
  iterator end();
  std::string value() const;
};