单链表 [英] Single linked list
问题描述
我已经创建了一个链表。一切都很好
I have created a single linked list. Everything works fine.
我只想知道我的代码中是否有任何潜在的危险。我关心的代码片段是我的推送，流行和清理。代码的部分只是为了用户交互，所以不是很重要（我发布了，所以我在做什么更清楚）。只是链表应用程序。
I just want to know if I have done anything potentially dangerous in my code. The code snippets I am concerned about is my push, pop, and clean-up. The parts of the code is just for user interaction so not really important (I posted anyway so that it was more clear in what I was doing). Just the linked list application.
非常感谢任何建议，因为这是我的拳头尝试。
Many thanks for any suggestions, as this is my fist attempt.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct product_data product_data_t;
struct product_data
{
int product_code;
char product_name[128];
int product_cost;
product_data_t *next;
};
static product_data_t *head = NULL;
static product_data_t *tail = NULL;
static product_data_t *new_product = NULL;
// Push a product on to the list.
void push(int code, char name[], int cost);
// Pop (delete) a product from the list.
void pop(int code);
// Display all product in the list.
void display_list();
// Delete all memory allocated on the list
void clean_up();
// Display menu
void menu();
int main(void)
{
menu();
getchar();
return 0;
}
void push(int code, char name[], int cost)
{
// Allocate memory for the new product
new_product = calloc(1, sizeof(product_data_t));
if(!new_product)
{
fprintf(stderr, "Cannot allocated memory");
exit(1);
}
/* Populate new products elements fields */
new_product->product_code = code;
strncpy(new_product->product_name, name, sizeof(new_product->product_name));
new_product->product_cost = cost;
new_product->next = NULL;
// Set the head and tail of the linked list
if(head == NULL)
{
// First and only product
head = new_product;
}
else
{
tail->next = new_product;
}
tail = new_product;
}
// Find the product by code and delete
void pop(int code)
{
product_data_t *product = head;
product_data_t *temp = NULL;
product_data_t *previous = head;
int found = 0; // 0 - Not Found, 1 - Found
if(!head)
{
puts("The list is empty");
return;
}
while(product)
{
if(product->product_code == code)
{
found = 1; // Found
// Check if this is in the first node - deleting from head
if(head->product_code == code)
{
temp = head;
head = head->next;
free(temp);
// Finished Deleting product
return;
}
// Check if this is the end node - deleting from the tail
if(tail->product_code == code)
{
temp = tail;
tail = previous;
free(temp);
// Finished deleting product
return;
}
// delete from list if not a head or tail
temp = product;
previous->next = product->next;
free(temp);
// Finished deleting product
return;
}
// Get the address of the previous pointer.
previous = product;
product = product->next;
}
if(!found)
{
printf("code [ %d ] was not found\n", code);
}
// Set all to null after finished with them
product = NULL;
temp = NULL;
previous = NULL;
}
// Traverse the linked list
void display_list()
{
// Start at the beginning
product_data_t *product = head;
while(product)
{
printf("===================================\n");
printf("Product code: \t\t%d\n", product->product_code);
printf("Product name: \t\t%s\n", product->product_name);
printf("product cost (USD): \t%d\n", product->product_cost);
printf("===================================\n\n");
// Point to the next product
product = product->next;
}
// Finished set to null
product = NULL;
}
// Release all resources
void clean_up()
{
product_data_t *temp = NULL;
while(head)
{
temp = head;
head = head->next;
free(temp);
}
head = NULL;
temp = NULL;
// End program - goodbye
exit(0);
}
void menu()
{
int choice = 0, code = 0, cost = 0;
char name[128] = {0};
do
{
fflush(stdin); // Flush the input buffer
puts("========= Welecome to linked list ===============");
puts("[1] Add new product to the list");
puts("[2] Delete a product from the list");
puts("[3] Display all products");
puts("[4] Exit and clean up");
printf("Enter your choice: ");
scanf("%d", &choice);
switch(choice)
{
case 1:
printf("Enter product code: ");
scanf("%d", &code);
printf("Enter cost: ");
scanf("%d", &cost);
printf("Enter name: ");
scanf("%s", name);
push(code, name, cost);
break;
case 2:
printf("Enter product code: ");
scanf("%d", &code);
pop(code);
break;
case 3:
display_list();
break;
case 4:
clean_up();
break;
default:
puts("Incorrect choice");
break;
}
}while(choice != 4);
}
推荐答案
从pop（） p>
From pop()
if(head->product_code == code)
{
temp = head;
head = head->next;
free(temp);
// Finished Deleting product
return;
}
在只有一个项目的情况下，'头'和'尾'将指向同一个节点。然而，如果你弹出这个项目，'head'将被调整，但'tail'仍将指向free'd节点。这将留下一个不好的指针，这可能会导致您的计算机爆炸。
In the case of there only being one item, 'head' and 'tail' would be pointing to the same node. However, if you pop this one item, 'head' will be adjusted but 'tail' will still be pointing to the free'd node. This will leave a bad pointer, which may cause your computer to explode.
附录：同样，如果您有new_product弹出被推送的最后一个节点，而clean_up（）将使'tail'指针悬空。即使提供的代码示例在它们空闲之后永远不会被引用，C代码中的悬挂指针应该被视为潜在危险。
Addendum: Similarly, 'new_product' will be dangling if you ever pop the last node that was pushed, and clean_up() will leave the 'tail' pointer dangling as well. Even if the code sample provided will never dereference these after they're free'd, dangling pointers in C code should always be treated as "potentially dangerous".
这篇关于单链表的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
