好的,所以我有一个用C编写的散列表,我使用单独的链接(链表)来解决冲突。我注意到,如果没有冲突,并且每个项都散列到自己的索引中,我可以释放整个表。但如果发生冲突,并且索引处有多个值,则只能释放第一个值,而不能释放该索引中的其余值。当程序试图释放该索引处的其他索引时,程序崩溃。我试着调试它,发现其他的值都被设置为空,我不知道为什么,因为当我将它们插入到表中时,我使用的是malloc。我知道我错过了什么。如果有人能帮忙的话,那就太棒了,因为我已经试着解决这个问题好几个小时了:/
代码如下:
int symTabSearch(struct hashTable * h, char * label);
int insertToSymTab(struct hashTable * h, char * label, int locctr);
struct listNode
{
char * label;
int address;
struct listNode * next;
};
struct hashTableNode
{
int blockCount; //number of elements in a block
struct listNode * firstNode;
};
struct hashTable
{
int tableSize;
int count; //number of elements in the table
struct hashTableNode * table;
};
struct hashTable * createHashTable(int size)
{
struct hashTable * ht;
ht = (struct hashTable*)malloc(sizeof(struct hashTable));
if (!ht)
return NULL;
ht->tableSize = size;
ht->count = 0;
ht->table = (struct hashTableNode *)malloc(sizeof(struct hashTableNode) * ht->tableSize);
if (!ht->table)
{
printf("Memory error\n");
return NULL;
}
int i;
for (i = 0; i < ht->tableSize; i++)
{
ht->table[i].blockCount = 0;
ht->table[i].firstNode = NULL;
}
return ht;
}
/*hash function: adds up the ascii values of each
character, multiplies by a prime number (37) and mods the sum wih the table size*/
int hash(char * label, int tableSize)
{
int hashVal = 0;
size_t i;
for (i = 0; i < strlen(label); i++)
hashVal = 37 * hashVal + label[i];
hashVal %= tableSize;
if (hashVal < 0)
hashVal += tableSize;
return hashVal;
}
int symTabSearch(struct hashTable * h, char * label)
{
struct listNode * temp;
temp = h->table[hash(label, h->tableSize)].firstNode; //temp points to the first listNode in table[hashedIndex]
while (temp)
{
if (strcmp(temp->label, label) == 0)
return 1; //found
temp = temp->next; //go to next link
}
return 0; //not found
}
int insertToSymTab(struct hashTable * h, char * label, int locctr)
{
int index;
struct listNode * currentNode, *newNode;
index = hash(label, h->tableSize);
currentNode = h->table[index].firstNode;
newNode = (struct listNode *)malloc(sizeof(struct listNode));
newNode->label = (char *)malloc(sizeof(char) * 7); //allocates 7 chars to store label up to 6 chars long (0-5), last one is for the '\0'
if (!newNode) //if new node is null
{
printf("Error creating new node\n");
return 0;
}
strcpy(newNode->label, label);
newNode->address = locctr;
if (h->table[index].firstNode == NULL) //if first node at table index is empty
{
h->table[index].firstNode = newNode;
h->table[index].firstNode->next = NULL;
}
else
{ //firstNode was not empty, so chain newNode to the next empty node
while (currentNode != NULL) //go to next available node
currentNode = currentNode->next;
currentNode = newNode;
currentNode->next = NULL;
}
h->table[index].blockCount++;
h->count++;
return 1;
}
void freeHashTable(struct hashTable * h) //might not free memory properly, might crash too, test later
{
int i, j;
struct listNode * current, *temp;
char * tempStr;
if (!h) //make sure table even has memory to be freed
return;
for (i = 0; i < h->tableSize; i++)
{
current = h->table[i].firstNode;
for (j = 0; j < h->table[i].blockCount; j++)
{
temp = current;
tempStr = current->label;
current = current->next;
free(temp);
free(tempStr);
temp = NULL;
tempStr = NULL;
}
}
free(h->table);
h->table = NULL;
free(h);
h = NULL;
}
最佳答案
当您试图将节点追加到列表中时,问题出在insertToSymTab
函数中。
这里的问题是这个循环:
while (currentNode != NULL) //go to next available node
currentNode = currentNode->next;
当循环完成时,您已经传递了列表的末尾,
currentNode
的值为NULL
。更改该指针不会将新节点追加到列表的末尾。相反,你需要改变你的循环。
while (currentNode->next != NULL)
currentNode = currentNode->next;
然后当循环结束时,
currentNode
将成为列表中当前的最后一个节点,您可以通过更改currentNode->next
来追加新节点:currentNode->next = newNode;
不要忘记将
newNode->next
设置为NULL
。关于c - 当C中发生冲突(使用单独的链接)时,无法释放哈希表中的节点,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/33361834/