Display()重置整个数组C++

Question

我尝试创建一个哈希映射，使用分离链接将其输入存储为链表节点。第一个显示函数提供了理想的输出，但下一个函数将整个数组重置为空nullptr。我使用了相同的类对象，所以它不应该每次都给出相同的结果吗？不知怎么的是因为析构函数？我想这可能是因为我插入了一个新项目，所以我删除了它，但同样的事情仍然存在。我唯一怀疑的是display((函数，但请指出问题是否来自其他地方。

很抱歉，如果帖子太长，我想确保每个人都能看到整个代码来发现问题。

#include <iostream>
#include <string>
#include "C:Usersadminsourcereposhash-library-mastersha3.cpp"
using namespace std;
const int TABLE_SIZE = 11;
struct HashNode
{
string key;
string value;
HashNode* next;
};
class HashMap {
private:
HashNode **table;
public:
//each element of table will be a root pointer to their respective chain
HashMap() {
table = new HashNode*[TABLE_SIZE];
for (int i = 0; i < TABLE_SIZE; i++)
{
table[i] = nullptr;
}   
}
//hashing algorithm using SHA3 (courtesy of Stephan Brumme)
string hashFunc(string input)
{
string key;
SHA3 sha3;
key = sha3(input);
return key;
}
//insert new node
void insert(string key, string value)
{
//using hashing function to calculate hash index from string variable key
int hash = 0;
for (int a = 0; a < key.length(); ++a)
hash += key[a];
hash = hash % TABLE_SIZE;
//create new node to store data
HashNode* newNode = new HashNode;
newNode->value = value;
newNode->key = key;
newNode->next = nullptr;
//check and insert new node to front of line
if (table[hash] == nullptr)
table[hash] = newNode;
else
{
newNode->next = table[hash]->next;
table[hash]->next = newNode;
}
}
void display()
{
for (int i = 0; i < TABLE_SIZE; ++i)
{
if (table[i] == nullptr)
cout << i << " NULL" << endl;
else
{
while (table[i] != nullptr)
{
cout << i << " " << table[i]->value << "; ";
table[i] = table[i]->next;
if (table[i] == nullptr)
cout << "(end of chain)" << endl;
}
}
}   
}
~HashMap() {
for (int i = 0; i < TABLE_SIZE; i++)
if (table[i] != NULL)
delete table[i];
delete[] table;
}
};

驱动

#include <iostream>
#include <string>
#include "hashMap.h"
using namespace std;
int main()
{
HashMap obj;

//test insert
obj.insert("5", "3100 Main St, Houston TX ");
obj.insert("5", "2200 Hayes Rd, Austin TX");
obj.insert("226", "1775 West Airport St, San Antonio TX");
obj.insert("273", "3322 Walnut Bend, Houston TX");
obj.insert("491", "5778 Alabama, Waco TX");
obj.insert("94", "3333 New St, Paris TX");
obj.display(); //resolved
cout << endl << endl;
//testing new hashing algorithm
string input, key;
cout << "Please enter any new address you want to store: ";
cin >> input;
key = obj.hashFunc(input); //create hash key
obj.insert(key, input);
obj.display(); //resets the array somehow
return 0;
}

输出

0 1775 West Airport St, San Antonio TX; (end of chain)
1 NULL
2 3322 Walnut Bend, Houston TX; (end of chain)
3 NULL
4 5778 Alabama, Waco TX; (end of chain)
5 NULL
6 NULL
7 NULL
8 NULL
9 3100 Main St, Houston TX ; 9 2200 Hayes Rd, Austin TX; (end of chain)
10 3333 New St, Paris TX; (end of chain)
//where display() resets
Please enter any new address you want to store: ewrewrw
0 NULL
1 ewrewrw; (end of chain)
2 NULL
3 NULL
4 NULL
5 NULL
6 NULL
7 NULL
8 NULL
9 NULL
10 NULL

Answer 1

将不应该更改对象本身的所有成员函数设为const。这样可以确保在const上下文中使用对象时可以使用该函数，并且如果您犯了错误，编译器也可以帮助您。如果您尝试修改函数中的对象，它会给您带来编译错误，因此会抱怨您在当前代码中对对象进行更改的table[i] = table[i]->next;行。

因此，从生成函数const开始，并修复错误。有了修复程序，它可能看起来像这样：

void display() const  // const added
{
for (int i = 0; i < TABLE_SIZE; ++i)
{
if (table[i] == nullptr)
cout << i << " NULL" << endl;
else
{
// using a temporary pointer, ptr, to go through the list
for(HashNode* ptr = table[i]; ptr != nullptr; ptr = ptr->next)
{
cout << i << " " << ptr->value << "; ";
}
cout << "(end of chain)" << endl;
}
}   
}

Answer 2

display函数通过循环将table的所有元素设置为nullptr，直到它们变为nullptr。

您应该使用另一个指针变量进行迭代，以避免这种破坏。

void display()
{
for (int i = 0; i < TABLE_SIZE; ++i)
{
if (table[i] == nullptr)
cout << i << " NULL" << endl;
else
{
HashNode *p = table[i]; // another pointer variable for iterating
while (p != nullptr)
{
cout << i << " " << p->value << "; ";
p = p->next;
if (p == nullptr)
cout << "(end of chain)" << endl;
}
}
}   
}