这里是代码,我用一个例子来运行它,但是当它出现时
比较我不明白什么是错的? ,提前感谢任何帮助。我需要正确打印字典文本(插入,打印),还不能提出解决方案,我的意思是使用字典数据结构。
#include <stdlib.h>
#include <ctype.h>
#include <string.h>
#include <assert.h>
#include <stdio.h>
typedef struct Node_s {
char *element;
struct Node_s *left, *right;
} Node;
typedef struct {
Node *head;
} Table;
//Table *initialize();
//Node *createNode(const char *element);
Table *initialize() {
Table *tb = malloc(sizeof(Table)*1000);
tb->head = NULL;
return tb;
}
Node *createNode( char * element ) {
Node *temp = malloc(sizeof(temp));
temp->element = element ;
temp->left = temp->right = NULL;
return temp;
}
void insert(Table *temp, char *element) {
Node *nd = createNode(element);
Table * place = NULL;
Node *new = NULL;
int cmp = 0;
if(temp->head == NULL) {
temp->head= nd;
printf("empty ! \n");
return;
}
else {
Table *current = temp;
while (current!=NULL) {
cmp = strcmp(current->head->element,element);
if(cmp < 0) {
current->head= current->head->left;
}
else if(cmp > 0) {
current->head = current->head->right;
}
} //while
place = current;
new = nd;
if(cmp > 0 ) {
place->head->right = new ;
}
else if(cmp <0 ) {
place->head->left = new;
}
}
}
void print_table(Table *temp) {
if(temp!=NULL || !temp->head) return;
print_table(temp->head->left);
printf("%s \n",temp->head->element);
print_table(temp->head->right);
}
int main () {
Node * nd = NULL;
//nd->element = "key";
// nd = createNode("key");
Table *tb = initialize();
//tb->head = createNode("key");
//tb->head = createNode("key");
insert(tb, "table element1");
insert(tb, "table element2");
insert(tb, "table element2");
//nd = createNode("key1");
// print_table(t);
//printf("%s \n",nd->element);
print_table(tb);
// printf("%s \n",tb->head->element);
free(nd);
return 0;
}
答案 0 :(得分:1)
这里存在很多潜在的错误,但您的主要问题在于createNode的以下行:
Node *temp = malloc(sizeof(temp));
这里你正在做一个sizeof(temp),temp是一个指针。这意味着您只为指针分配足够的内存(通常为8个字节)。因此,在使用堆分配结构的左/右成员时,您正在编写分配内存之外的内容。修复:
Node *temp = malloc(sizeof(Node));
// EXTRA: I also recommend that you verify that the allocation was successful
if (temp) {
temp->element = element ;
temp->left = temp->right = NULL;
}
return temp;
在printTable中,您还应验证temp本身不是NULL,因为您传递的函数参数可能为NULL:
if(!temp || !temp->head) return;
此外,删除main末尾的free(nd);
,因为在未分配的堆内存上调用free()会破坏堆,并且通常会导致段错误。
答案 1 :(得分:1)
您的打印方法在到达左侧的最后一个节点时崩溃,因为它会调用print_table(NULL),因为左侧没有其他内容。之后,当它执行行
if(!temp->head) return;
由于temp为NULL,您会收到内存访问冲突,还应检查temp本身是否为NULL。
if( !temp || !temp->head ) return;
那应该解决你的问题。
答案 2 :(得分:1)
第二个问题是第二次拨打insert
:
while (current != NULL) {
cmp = strcmp(current->head->element, element); // this line
您没有检查current->head
本身是否为NULL。根据您实施的内容,您使用head
作为哨兵,因此它可以为NULL。但是,您的搜索循环完全忘记了这种情况,并假设head
永远不会为NULL。
你的循环看起来并不正确。你遍历左边,所以如果左分支&#34;用完了将会发生的事情。 (就像你第二次打电话给insert
时那样)?
此外,您的insert
函数存在内存泄漏。您可能在此处分配2个新节点:
Node *nd = createNode(element);
在这里:
new = createNode(element);
只有一个存储而另一个存在泄漏。
另一个问题是,如果两个项目相同,那么您的树在while
循环中不执行任何操作。两个相等的项会导致无限循环:
while (current!=NULL)
{
cmp = strcmp(current->head->element,element);
if(cmp < 0)
current->head= current->head->left;
else if(cmp > 0)
current->head = current->head->right;
else
printf("these are equal ! \n"); // but we don't do anything with current!
}
如果目标是没有重复项,那么如果找到重复项,则应退出此函数。如果目标是存储重复项,则只测试< 0
,其他任何内容都会在正确的分支上进行。
答案 3 :(得分:0)
This might be what you are looking for.
It handles a doubly linked list
error checking is added
removed undesirable/unnecessary typedef's from struct definitions
corrected the logic to link in new nodes
avoided recursion in the printing of the linked list
#include <stdlib.h>
#include <ctype.h>
#include <string.h>
#include <assert.h>
#include <stdio.h>
struct Node
{
char *element;
struct Node *left;
struct Node *right;
};
// define the head pointer for the linked list
struct Node *head = NULL;
// struct Node *createNode(const char *element);
struct Node *createNode( char * element )
{
struct Node *pNewNode = NULL;
if( NULL == (pNewNode = malloc(sizeof(struct Node)) ) )
{ // then, malloc failed
perror( "malloc for new node failed" );
exit( EXIT_FAILURE );
}
// implied else, malloc successful
pNewNode->element = element ; // copies a char pointer
pNewNode->left = NULL;
pNewNode->right = NULL;
return pNewNode;
} // end function: createNode
void insert(char *element)
{
int cmp = 0;
// get ptr to first node in list
struct Node *pCurrentNode = head;
// create the node to be inserted into linked list
struct Node *pNewNode = createNode(element);
if (pCurrentNode == NULL)
{ // then list empty
head = pNewNode;
printf("added first node\n");
return;
}
// implied else, not first node
while (pCurrentNode->right)
{
cmp = strcmp(pCurrentNode->element,element);
if(cmp < 0)
{
// insert new node before current node
pNewNode->right = pCurrentNode;
pNewNode->left = pCurrentNode->left;
pCurrentNode->left = pNewNode;
(pNewNode->left)->right = pNewNode;
}
else if(cmp > 0)
{
// step to next node
pCurrentNode = pCurrentNode->right;
} // end if
// note: if data same, don't insert new node
} //while
if( pCurrentNode->right == NULL )
{ // then, reached end of list
// append new node to end of list
pNewNode->left = pCurrentNode;
pNewNode->right = NULL;
pCurrentNode->right = pNewNode;
} // end if
} // end function: insert
void print_table()
{
struct Node *pCurrentNode = head;
if( pCurrentNode == NULL ) return;
// implied else, list not empty
while( pCurrentNode )
{
printf("%s \n",pCurrentNode->element);
pCurrentNode = pCurrentNode->right;
} // end while
} // end function: print_table
void cleanup()
{
struct Node *pCurrentNode = head;
while( pCurrentNode )
{
pCurrentNode = pCurrentNode->right;
free( pCurrentNode->left );
}
} // end function: cleanup
int main ()
{
// exercise the insert function
insert("table element1"); // append first element
insert("table element2"); // append second element
insert("table element4"); // append third element
insert("table element3"); // insert forth element
insert("table element3"); // duplicate within list
insert("table element4"); // duplicate at end of list
print_table();
cleanup();
return 0;
} // end function: main
答案 4 :(得分:0)
我尝试了不同的实现,它编译并运行,它不允许重复。
#include <stdlib.h>
#include <ctype.h>
#include <string.h>
#include <stdio.h>
#include <assert.h>
#define ELEMENT_SIZE 1024
typedef struct Node_s
{
char element[ELEMENT_SIZE];
struct Node_s *left, *right;
} Node;
Node * createNode(char *element)
{
Node *node = malloc(sizeof(Node));
node->left = NULL;
node->right = NULL;
memcpy(node->element, element, ELEMENT_SIZE);
return node;
}
void free_node(Node *node)
{
if(!node)
return;
free_node(node->left);
free_node(node->right);
free(node);
}
Node * insert(Node **head_ptr, char *element)
{
Node *head = *head_ptr;
if(head == NULL){
Node *node = createNode(element);
head = node;
*head_ptr = node;
return node;
}else{
int comp = strcmp(head->element, element);
if(comp < 0){
// go left
if(head->left == NULL){
// set element to be temp left
Node *node = createNode(element);
head->left = node;
return node;
}else{
return insert(&head->left, element);
}
}else if(comp > 0){
// go right
if(head->right == NULL){
// set element to be temp left
Node *node = createNode(element);
head->right = node;
return node;
}else{
return insert(&head->right, element);
}
}else{
// element exists
printf("Element \"%s\" already exists\n", element);
return NULL;
}
}
}
void print_table(Node *temp)
{
if(!temp)
return;
printf("%s \n",temp->element);
print_table(temp->left);
print_table(temp->right);
}
int main ()
{
Node *nd = NULL;
printf("Address of nd is %p\n", &nd);
Node *n1 = insert(&nd, "table element 1");
n1 = insert(&nd, "table element 2");
n1 = insert(&nd, "table element 3");
n1 = insert(&nd, "element 1");
n1 = insert(&nd, "element 2");
n1 = insert(&nd, "element 3");
n1 = insert(&nd, "alternative 1");
n1 = insert(&nd, "alternative 2");
n1 = insert(&nd, "alternative 3");
n1 = insert(&nd, "alternative 1");
n1 = insert(&nd, "alternative 2");
n1 = insert(&nd, "alternative 3");
print_table(nd);
free_node(nd);
return 0;
}