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奇怪的循环模板模式-不能创建多个派生类吗?

static-polymorphism c++14 inheritance templates c++

0

KeyC0de · 技术社区 · 7 年前

#include <iostream>

static int nodes = 0;

class TreeNode {
private:
    int m_id;
public:
    TreeNode() : 
        m_id(++nodes)
    {}
    TreeNode(int id) :
        m_id(id)
    {
        ++nodes;
    }
    TreeNode* left;
    TreeNode* right;

    int getId() const {
        return m_id;
    }
};


template<typename T>
//typename std::enable_if<std::is_base_of<TreeParser, T>::value>::type
class TreeParser {
protected:
    TreeParser() {
        ++parsers;
    }
public:
    static uint32_t parsers;
    void preorderTraversal(TreeNode* node) {
        if (node != nullptr) {
            processNode(node);
            preorderTraversal(node->left);
            preorderTraversal(node->right);
        }
    }
    virtual ~TreeParser() = default;

    void processNode(TreeNode* node) {              // 2, 3. the generic algorithm is customized by derived classes
        static_cast<T*>(this)->processNode(node);   // depending on the client's demand - the right function will be called
    }
};

template<class T>
uint32_t TreeParser<T>::parsers = 0;

class SpecializedTreeParser1 : public TreeParser<SpecializedTreeParser1> // 1. is-a relationship
{
public:
    explicit SpecializedTreeParser1() : 
        TreeParser()
    {}
    void processNode(TreeNode* node) {
        std::cout << "Customized (derived - SpecializedTreeParser1) processNode(node) - "
            "id=" << node->getId() << '\n';
    }
};

class SpecializedTreeParser2 : public TreeParser<SpecializedTreeParser2> // 1. is-a relationship
{
public:
    explicit SpecializedTreeParser2() : 
        TreeParser()
    {}
    void processNode(TreeNode* node) {
        std::cout << "Customized (derived - SpecializedTreeParser2) processNode(node) - "
            "id=" << node->getId() << '\n';
    }
};


int main() 
{
    TreeNode root;
    TreeNode leftChild;
    TreeNode rightChild;

    root.left = &leftChild;
    root.right = &rightChild;

    std::cout << "Root id: " << root.getId() << '\n';
    std::cout << "Left child id: " << leftChild.getId() << '\n';
    std::cout << "Right child id: " << rightChild.getId() << '\n';

    SpecializedTreeParser1 _1;
    _1.preorderTraversal(&root);

    SpecializedTreeParser2 _2;
    _2.preorderTraversal(&root);
}

输出为:

Root id: 1
Left child id: 2
Right child id: 3
Customized (derived - SpecializedTreeParser1) preorderTraversal() - id=1
Customized (derived - SpecializedTreeParser1) preorderTraversal() - id=2
Customized (derived - SpecializedTreeParser1) preorderTraversal() - id=1963060099 // what is that?

2 回复 | 直到 7 年前

1

5

Michael Kenzel 7 年前

left 和 right 指针 leftChild 和 rightChild preorderTraversal() 到了那里,你的程序就崩溃了。这也是为什么你会觉得奇怪 id 最后:它是从某个随机记忆位置读取的

要解决此问题,请确保 左边 正确的 组织的成员 TreeNode 总是初始化为 nullptr 正如代码的其余部分所期望的:

class TreeNode {
    â¦
    TreeNode* left = nullptr;
    TreeNode* right = nullptr;
    â¦
};

2

3

bolov 7 年前

问题是 SpecializedTreeParser2 _2; TreeParser leftChild 和 rightChild 很好,他们的 id 已打印。检查输出

不, 和 右孩子 初始化不正确,问题不存在 SpecializedTreeParser2 _2

TreeNode::left; 和 TreeNode* right; 美丽不确定的行为是什么 整个计划 具有未定义的行为,而不仅仅是使用未初始化的变量。这意味着尽管 SpecializedTreeParser1 _1 _1 似乎在工作 _2 不是。这是未定义的行为。修正它,不要试图理解为什么会有这种特殊的行为。

int main()
{
    int a = 24;
    std::cout << a << std::endl;

    int b; // uninitialized
    std::cout << b << std::endl; // <-- this line makes the WHOLE program to have UB
}

那么上述程序可能的行为是什么呢?

print "24"
print <garbage>

是可能的,也是合理的

print "24"
print "0"

也有可能

print "24"
<crash>

也不奇怪

print <garbage>
print "24"

也有可能。。。等等什么?对!!程序的未定义性质不必在行中显示 std::cout << b << std::endl . 程序的整个执行过程是未定义的。 有什么事吗 可能发生!

所有这些都是可能的:

<crash>

print "0"
print "0"

print "24"
print "24"

print "24"
print "0"
<crash>

print "Here be dragons"