问题描述
我正在使用C语言编写一个程序，在该程序中我必须使用链表，并且在该程序中，如果用户传递了场所的值，则必须在链表的开头插入新节点0，并且如果用户在 choice 变量中传递了位置1的值，则还将新节点插入到链表的末尾。但是我没有在控制台上获得任何输出，我的程序仅通过编写 Output 结束 我找不到我的代码中的问题，这是我的代码。
/*
program for making nodes and adding them in memory as per 0
and 1
0 means that insert the number at front,in other words insert number after head
1 means insert number at the last place
First you need to input a number and then enter the place you want to insert it by giving input as 0 and 1
*Recall what does 0 and 1 mean by looking at line 5-7 respectively.
Just like
5 0 6 1 7 0 8 1
*/
#include <stdio.h>
#include <stdlib.h>
// declaring struct with typedef for ease of use
typedef struct node
{
int data;
struct node *next;
}node;
// declarations of functions use for this program respectively
void free_node(struct node *head);
void insert_at_beg(int num,struct node *head);
void insert_at_end(int num,struct node *head);
void print_node(struct node *head);
int main(void)
{
struct node *head = NULL;
int n;
// taking input
printf("Input number of nodes: ");
scanf("%d",&n);
int num,choice;
printf("\nInput data for nodes->\n");
// loop which takes value and choice
for (int i = 0; i < n; i++)
{
num = 0,choice = 0;
printf("\nInput data for the %d node: ",i+1);
scanf("%d",&num);
do
{
printf("Input place for the %d node: ",i+1);
scanf("%d",&choice);
}
while (choice != 1 && choice != 0);
if (choice == 0)
{
// function to insert node at front of head
insert_at_beg(choice,head);
}
else
{
// function to insert node at last place
insert_at_end(choice,head);
}
}
// function to print nodes
print_node(head);
// function to free memory made by malloc()
free_node(head);
}
// function to free the nodes
void free_node(struct node *head)
{
struct node *temp = head;
while(temp != NULL)
{
free(temp);
temp = temp->next;
}
}
// function for inserting number at front
void insert_at_beg(int num,struct node *head)
{
struct node *new_node = malloc(sizeof(node));
if (new_node == NULL)
{
printf("Can't allocate memory.");
exit (1);
}
new_node->data = num;
new_node->next = head;
head = new_node;
}
// function for inserting node at end
void insert_at_end(int num,struct node *head)
{
struct node *new_node,*last_node = NULL;
new_node = malloc(sizeof(node));
if (new_node == NULL)
{
printf("Can't allocate memory.");
exit (1);
}
if (head == NULL)
{
new_node->data = num;
new_node->next = NULL;
head = new_node;
}
last_node = head;
new_node->data = num;
new_node->next = NULL;
while (last_node->next != NULL)
{
last_node = last_node->next;
}
last_node->next = new_node;
}
//function for printing nodes
void print_node(struct node *head)
{
printf("\nOutput: \n");
struct node *temp = head;
while(temp != NULL)
{
printf("%d ",temp->data);
temp = temp->next;
}
}
解决方法
您需要通过引用函数insert_at_beg和insert_at_end并将指针传递到头节点，并且需要函数free_node进行引用。
在C中通过引用传递意味着间接通过对象的指针传递对象。
例如，函数insert_at_beg的外观如下。请注意，这样的功能不应发出任何消息。该函数的调用者决定是否输出消息。
// function for inserting number at front
int insert_at_beg( struct node **head,int num )
{
struct node *new_node = malloc( sizeof( struct node ) );
int success = new_node != NULL;
if ( success )
{
new_node->data = num;
new_node->next = *head;
*head = new_node;
}
return success;
}
相应地，函数insert_at_end的外观如下
// function for inserting node at end
int insert_at_end( struct node **head,int num )
{
struct node *new_node = malloc( sizeof( struct node ) );
int success = new_node != NULL;
if ( success )
{
new_node->data = num;
new_node->next = NULL;
while ( *head != NULL )
{
head = &( *head )->next;
}
*head = new_node;
}
return success;
}
函数free_node具有未定义的行为，因为您正在使用指针temp访问已经释放的内存。
free(temp);
temp = temp->next;
可以通过以下方式定义功能
// function to free the nodes
void free_node( struct node **head )
{
while( *head != NULL )
{
struct node *temp = *head;
head = &( *head )->next;
free( temp );
}
}
可以像这样调用函数
insert_at_end( &head,num );
或
if ( !insert_at_end( &head,num ) )
{
printf( "There is no enough memory to insert the value %d\n",num );
}
函数print_node的参数应带有限定符const，因为列表在函数中未更改
//function for printing nodes
void print_node( const struct node *head )
{
printf("\nOutput: \n");
const struct node *temp = head;
//...
C是传递值的语言-复制传递给函数的值，并且对函数中的参数所做的更改不会影响调用方。因此head永远不会成为非null； head中对insert_at_end的分配是本地的，不会更新main中的head指针。