C++ 调试断言失败！表达式：块类型是否有效（标题->；\u块使用）
程序正确运行并输出所有内容，并抛出此错误。对于这意味着什么，以及如何找到或修复它，我还没有找到任何好的解释。下面是代码的完整副本（非常难看而且写得很糟糕）：C++ 调试断言失败！表达式：块类型是否有效（标题->；\u块使用）,c++,C++,程序正确运行并输出所有内容，并抛出此错误。对于这意味着什么，以及如何找到或修复它，我还没有找到任何好的解释。下面是代码的完整副本（非常难看而且写得很糟糕）： #包括 使用名称空间std； /*用于存储二叉树的类*/ 模板 类二叉树{ 受保护的： 类型parentArray[10]； 类型childArray[10]； 公众： 二叉树（）； 二叉树（int&k）； ~BinaryTree（）； BinaryTree（BinaryTree&bt）； void运算符=（二进制树&bt）； friend
#包括
使用名称空间std；
/*用于存储二叉树的类*/
模板
类二叉树{
受保护的：
类型parentArray[10]；
类型childArray[10]；
公众：
二叉树（）；
二叉树（int&k）；
~BinaryTree（）；
BinaryTree（BinaryTree&bt）；
void运算符=（二进制树&bt）；
friend ostream&operator您的二进制树
析构函数始终确保：
#include <iostream>
using namespace std;
/* a class for storing a Binary Tree */
template <class Type>
class BinaryTree {
protected:
Type parentArray[10];
Type childArray[10];
public:
BinaryTree();
BinaryTree(int& k);
~BinaryTree();
BinaryTree(BinaryTree<Type>& bt);
void operator= (BinaryTree<Type>& bt);
friend ostream& operator<< (ostream& s, BinaryTree<Type> bt) {
s << "[ ";
bt.inorder(bt.getRoot());
s << "]" << endl;
return s;
};
int size();
int height();
int getLeft(int k);
int getRight(int k);
void preorder(int k);
void inorder(int k) {
// do I have a left child?
if ((getLeft(k)) != -1) {
// if yes inorder (left child)
inorder(getLeft(k));
};
// output k
cout << k << " ";
// do I have a right child?
if ((getRight(k)) != -1) {
// if yes inorder (right child)
inorder(getRight(k));
};
};
void postorder(int k);
void setRoot(Type& val);
void setParent(Type* child, Type* parent);
void setLeft(Type& val);
void setRight(Type& val);
int getRoot();
};
/* default constructor */
template <class Type>
BinaryTree<Type>::BinaryTree() {
parentArray = new ArrayClass<Type>();
childArray = new ArrayClass<Type>();
};
/* non-empty constructor */
template <class Type>
BinaryTree<Type>::BinaryTree(int& k) {
// parentArray = new Type[k];
// childArray = new Type[k];
};
template <class Type>
BinaryTree<Type>::~BinaryTree() {
delete[] parentArray;
delete[] childArray;
};
template <class Type>
BinaryTree<Type>::BinaryTree(BinaryTree<Type>& bt) {
for (int i = 0; i < bt.size(); i++) {
parentArray[i] = bt.parentArray[i];
childArray[i] = bt.childArray[i];
};
};
template <class Type>
void BinaryTree<Type>::operator= (BinaryTree<Type>& bt) {
};
/* return the size of the tree using the length of the parent array */
template <class Type>
int BinaryTree<Type>::size() {
return (sizeof(parentArray)/sizeof(*parentArray));
};
template <class Type>
int BinaryTree<Type>::height() {
return 5;
};
template <class Type>
int BinaryTree<Type>::getLeft(int k) {
// if the parent array value of the given number is k and
// the child array value indicates it is a left child
for (int i = 0; i < size(); i++) {
if ((parentArray[i] == k) && (childArray[i] == 0)) {
// return that value
return i;
};
};
return -1;
};
template <class Type>
int BinaryTree<Type>::getRight(int k) {
// if the parent array value of the given number is k and
// the child array value indicates it is a right child
for (int i = 0; i < size(); i++) {
if ((parentArray[i] == k) && (childArray[i] == 1)) {
// return that value
return i;
};
};
return -1;
};
template <class Type>
void BinaryTree<Type>::preorder(int k) {
// output k
cout << k << " ";
// do I have a left child?
if ((getLeft(k)) != -1) {
// if yes preorder left child
preorder(getLeft(k));
};
// do I have a right child?
if ((getRight(k)) != -1) {
// if yes preorder right child
preorder(getRight(k));
};
};
template <class Type>
void BinaryTree<Type>::postorder(int k) {
// do I have a left child?
if ((getLeft(k)) != -1) {
// if yes inorder (left child)
inorder(getLeft(k));
};
// do I have a right child?
if ((getRight(k)) != -1) {
// if yes inorder (right child)
inorder(getRight(k));
};
// output k
cout << k << " ";
};
template <class Type>
void BinaryTree<Type>::setRoot(Type& val) {
// if the given value is the root of the tree then set
// its index in the parent and child arrays to -1
parentArray[val] = -1;
childArray[val] = -1;
};
template <class Type>
void BinaryTree<Type>::setParent(Type* child, Type* parent) {
// set a given value as the parent of a given value
parentArray[(*child)] = *parent;
};
template <class Type>
void BinaryTree<Type>::setLeft(Type& val) {
// set a given value in the child array to indicate a left child
childArray[val] = 0;
};
template <class Type>
void BinaryTree<Type>::setRight(Type& val) {
// set a given value in the child array to indicate a right child
childArray[val] = 1;
};
template <class Type>
int BinaryTree<Type>::getRoot() {
// find the root value of the tree
for (int i = 0; i < size(); i++) {
if (parentArray[i] == -1) {
// and return it
return i;
};
};
};
int main() {
int* val1 = new int;
int* val2 = new int;
int* val3 = new int;
int count;
cin >> count;
BinaryTree<int> bt(count);
for (int i = 0; i < count; i++) {
cin >> *val1;
cin >> *val2;
cin >> *val3;
if (i == 0) {
bt.setRoot(*val1);
};
if (*val2 != -1) {
bt.setParent(val2, val1);
bt.setLeft(*val2);
}
if (*val3 != -1) {
bt.setParent(val3, val1);
bt.setRight(*val3);
}
val1 = new int;
val2 = new int;
val3 = new int;
};
cout << bt.size() << endl;
bt.postorder(bt.getRoot());
cout << endl;
bt.preorder(bt.getRoot());
cout << endl;
delete val1;
delete val2;
delete val3;
};
不幸的是，该类的一个构造函数没有new
这些数组中的任何一个。因此，析构函数最终尝试删除一对未初始化的垃圾指针
也有可能是这个类，但我还没有充分分析
编辑：正如在评论中指出的，这些不是指针；因此这是错误的，但出于其他原因。这些数组不是指针。不需要新建它们。它们不应该被删除。主中的for
循环正在泄漏内存。val1
，val2
val3是循环底部的新代码，但从未释放。建议：在测试之前更频繁地测试或编写较小的代码块。当测试一个函数的新代码时，通常很容易找出错误所在。编写构造函数和析构函数并进行测试。添加一个方法。测试。添加另一个方法并进行测试。冲洗。重复.这回答了你的问题吗？
#include <iostream>
using namespace std;
/* a class for storing a Binary Tree */
template <class Type>
class BinaryTree {
protected:
Type parentArray[10];
Type childArray[10];
public:
BinaryTree();
BinaryTree(int& k);
~BinaryTree();
BinaryTree(BinaryTree<Type>& bt);
void operator= (BinaryTree<Type>& bt);
friend ostream& operator<< (ostream& s, BinaryTree<Type> bt) {
s << "[ ";
bt.inorder(bt.getRoot());
s << "]" << endl;
return s;
};
int size();
int height();
int getLeft(int k);
int getRight(int k);
void preorder(int k);
void inorder(int k) {
// do I have a left child?
if ((getLeft(k)) != -1) {
// if yes inorder (left child)
inorder(getLeft(k));
};
// output k
cout << k << " ";
// do I have a right child?
if ((getRight(k)) != -1) {
// if yes inorder (right child)
inorder(getRight(k));
};
};
void postorder(int k);
void setRoot(Type& val);
void setParent(Type* child, Type* parent);
void setLeft(Type& val);
void setRight(Type& val);
int getRoot();
};
/* default constructor */
template <class Type>
BinaryTree<Type>::BinaryTree() {
parentArray = new ArrayClass<Type>();
childArray = new ArrayClass<Type>();
};
/* non-empty constructor */
template <class Type>
BinaryTree<Type>::BinaryTree(int& k) {
// parentArray = new Type[k];
// childArray = new Type[k];
};
template <class Type>
BinaryTree<Type>::~BinaryTree() {
delete[] parentArray;
delete[] childArray;
};
template <class Type>
BinaryTree<Type>::BinaryTree(BinaryTree<Type>& bt) {
for (int i = 0; i < bt.size(); i++) {
parentArray[i] = bt.parentArray[i];
childArray[i] = bt.childArray[i];
};
};
template <class Type>
void BinaryTree<Type>::operator= (BinaryTree<Type>& bt) {
};
/* return the size of the tree using the length of the parent array */
template <class Type>
int BinaryTree<Type>::size() {
return (sizeof(parentArray)/sizeof(*parentArray));
};
template <class Type>
int BinaryTree<Type>::height() {
return 5;
};
template <class Type>
int BinaryTree<Type>::getLeft(int k) {
// if the parent array value of the given number is k and
// the child array value indicates it is a left child
for (int i = 0; i < size(); i++) {
if ((parentArray[i] == k) && (childArray[i] == 0)) {
// return that value
return i;
};
};
return -1;
};
template <class Type>
int BinaryTree<Type>::getRight(int k) {
// if the parent array value of the given number is k and
// the child array value indicates it is a right child
for (int i = 0; i < size(); i++) {
if ((parentArray[i] == k) && (childArray[i] == 1)) {
// return that value
return i;
};
};
return -1;
};
template <class Type>
void BinaryTree<Type>::preorder(int k) {
// output k
cout << k << " ";
// do I have a left child?
if ((getLeft(k)) != -1) {
// if yes preorder left child
preorder(getLeft(k));
};
// do I have a right child?
if ((getRight(k)) != -1) {
// if yes preorder right child
preorder(getRight(k));
};
};
template <class Type>
void BinaryTree<Type>::postorder(int k) {
// do I have a left child?
if ((getLeft(k)) != -1) {
// if yes inorder (left child)
inorder(getLeft(k));
};
// do I have a right child?
if ((getRight(k)) != -1) {
// if yes inorder (right child)
inorder(getRight(k));
};
// output k
cout << k << " ";
};
template <class Type>
void BinaryTree<Type>::setRoot(Type& val) {
// if the given value is the root of the tree then set
// its index in the parent and child arrays to -1
parentArray[val] = -1;
childArray[val] = -1;
};
template <class Type>
void BinaryTree<Type>::setParent(Type* child, Type* parent) {
// set a given value as the parent of a given value
parentArray[(*child)] = *parent;
};
template <class Type>
void BinaryTree<Type>::setLeft(Type& val) {
// set a given value in the child array to indicate a left child
childArray[val] = 0;
};
template <class Type>
void BinaryTree<Type>::setRight(Type& val) {
// set a given value in the child array to indicate a right child
childArray[val] = 1;
};
template <class Type>
int BinaryTree<Type>::getRoot() {
// find the root value of the tree
for (int i = 0; i < size(); i++) {
if (parentArray[i] == -1) {
// and return it
return i;
};
};
};
int main() {
int* val1 = new int;
int* val2 = new int;
int* val3 = new int;
int count;
cin >> count;
BinaryTree<int> bt(count);
for (int i = 0; i < count; i++) {
cin >> *val1;
cin >> *val2;
cin >> *val3;
if (i == 0) {
bt.setRoot(*val1);
};
if (*val2 != -1) {
bt.setParent(val2, val1);
bt.setLeft(*val2);
}
if (*val3 != -1) {
bt.setParent(val3, val1);
bt.setRight(*val3);
}
val1 = new int;
val2 = new int;
val3 = new int;
};
cout << bt.size() << endl;
bt.postorder(bt.getRoot());
cout << endl;
bt.preorder(bt.getRoot());
cout << endl;
delete val1;
delete val2;
delete val3;
};
delete[] parentArray;
delete[] childArray;