C++ C++；避免检查双指针持有的未初始化值

解决

问题最终源于数据结构的设计。要删除根元素，必须有一个堆分配的（新的）指针指向它，这在原始情况下是不可能的，因为数据直接由树保存。现在有一个

节点

结构保存所有数据，树类包含所有方法和根指针

---

我有一个双指针

target\u address

初始化为

NULL

，我以后必须检查

*target\u address

是否为NULL。在这两者之间，函数保证

*target_address

指向未初始化的值，但Valgrind一直抱怨我只有在删除根元素时才读取未初始化的值

树_delete也抱怨未初始化值；我敢肯定它也在抱怨*目标地址

void tree<T>::tree_delete(tree<T>** target_address) {
    tree<T>* target = *target_address;
    // first case: no left subtree, replace with right subtree (or NULL for a leaf) 
    if (!target->left) 
        *target_address = target->right;

    // second case: no right subtree, replace with left subtree
    else if (!target->right) 
        *target_address = target->left;

    // third case: both subtrees, find second largest by taking rightmost left tree
    else {
        tree<T>** second_largest_address = &target->left;  // start at target's left tree
        while ((*second_largest_address)->right)  // keep going right
            second_largest_address = &((*second_largest_address)->right);  
        // reached the rightmost left
        tree<T>* second_largest = *second_largest_address;
        *target_address = second_largest;  // delete target by replacing it with second largest
        // second largest guranteed to not have a right subtree, so can treat as case 2 by shifting
        *second_largest_address = second_largest->left;  
        second_largest->left = target->left;
        second_largest->right = target->right;
    }
    delete target;
}

类似于的代码，如果（！this）返回空非常可疑。这是一个无操作，并指示其他地方未定义的行为。难怪Valgrind报告您的代码出现问题

更糟糕的是，请查看

树中的代码\u find

：

tree<T>* root = this;   // <-- ensures *target_address points to stack variable !
tree<T>** target_address = &root; // Set out pointer to stack variable

---

tree*root=this；//能否显示
tree:：data
的定义？
tree\u find
如果数据与第一个节点匹配，则返回指向堆栈变量的指针
tree*root
是堆栈变量，
tree**target\u address=&root
-堆栈变量的地址，
*target\u address\u handle=target\u address-将不存在的变量的地址发送回调用方。我不太清楚为什么要返回指向所需节点的指针的地址，而不仅仅是指向所需节点的指针！树数据只是
T数据
，其他属性是
Tree*left，*right
typeid（T）==typeid（T）
那么总是正确的@MattMcNabb啊对，这证实了我的怀疑。。。想想看，我不认为我需要三个指针；我会尝试摆脱它，看看它是否仍然有效，是的，typeid（T）=typeid（T）。。。忘记C++是强类型的，可能有人可以尝试在不同类型的树上搜索
void tree<T>::tree_delete(tree<T>** target_address) {
    tree<T>* target = *target_address;
    // first case: no left subtree, replace with right subtree (or NULL for a leaf) 
    if (!target->left) 
        *target_address = target->right;

    // second case: no right subtree, replace with left subtree
    else if (!target->right) 
        *target_address = target->left;

    // third case: both subtrees, find second largest by taking rightmost left tree
    else {
        tree<T>** second_largest_address = &target->left;  // start at target's left tree
        while ((*second_largest_address)->right)  // keep going right
            second_largest_address = &((*second_largest_address)->right);  
        // reached the rightmost left
        tree<T>* second_largest = *second_largest_address;
        *target_address = second_largest;  // delete target by replacing it with second largest
        // second largest guranteed to not have a right subtree, so can treat as case 2 by shifting
        *second_largest_address = second_largest->left;  
        second_largest->left = target->left;
        second_largest->right = target->right;
    }
    delete target;
}

==25887== Conditional jump or move depends on uninitialised value(s)
==25887==    at 0x401764: tree<int>::remove(int) (in /home/johnson/Code/tree/treetest)
==25887==    by 0x4013AA: main (in /home/johnson/Code/tree/treetest)
==25887==  Uninitialised value was created by a stack allocation
==25887==    at 0x40175A: tree<int>::remove(int) (in /home/johnson/Code/tree/treetest)
==25887== 
After reading *target tree address
Before read
==25887== Use of uninitialised value of size 8
==25887==    at 0x401A75: tree<int>::tree_delete(tree<int>**) (in /home/johnson/Code/tree/treetest)
==25887==    by 0x40178E: tree<int>::remove(int) (in /home/johnson/Code/tree/treetest)
==25887==    by 0x4013AA: main (in /home/johnson/Code/tree/treetest)
==25887==  Uninitialised value was created by a stack allocation
==25887==    at 0x401A46: tree<int>::tree_delete(tree<int>**) (in /home/johnson/Code/tree/treetest)
==25887== 
==25887== Use of uninitialised value of size 8
==25887==    at 0x401AA5: tree<int>::tree_delete(tree<int>**) (in /home/johnson/Code/tree/treetest)
==25887==    by 0x40178E: tree<int>::remove(int) (in /home/johnson/Code/tree/treetest)
==25887==    by 0x4013AA: main (in /home/johnson/Code/tree/treetest)
==25887==  Uninitialised value was created by a stack allocation
==25887==    at 0x401A46: tree<int>::tree_delete(tree<int>**) (in /home/johnson/Code/tree/treetest)
==25887== 
==25887== Use of uninitialised value of size 8
==25887==    at 0x401AF2: tree<int>::tree_delete(tree<int>**) (in /home/johnson/Code/tree/treetest)
==25887==    by 0x40178E: tree<int>::remove(int) (in /home/johnson/Code/tree/treetest)
==25887==    by 0x4013AA: main (in /home/johnson/Code/tree/treetest)
==25887==  Uninitialised value was created by a stack allocation
==25887==    at 0x401A46: tree<int>::tree_delete(tree<int>**) (in /home/johnson/Code/tree/treetest)
==25887== 
==25887== Invalid read of size 8
==25887==    at 0x401AF5: tree<int>::tree_delete(tree<int>**) (in /home/johnson/Code/tree/treetest)
==25887==    by 0x40178E: tree<int>::remove(int) (in /home/johnson/Code/tree/treetest)
==25887==    by 0x4013AA: main (in /home/johnson/Code/tree/treetest)
==25887==  Address 0x801f0fc36d is not stack'd, malloc'd or (recently) free'd

tree<T>* root = this;   // <-- ensures *target_address points to stack variable !
tree<T>** target_address = &root; // Set out pointer to stack variable