链表中 char、char* 的区别

Question

我创建了一个包含 int 和 char 类型数据的链表。一个函数将数据添加到列表中，另一个函数将其打印出来。当我只打印 int 类型时，我没有遇到任何问题，但是当我尝试同时打印 char 类型时，程序崩溃了。

所以它必须按照我在打印函数 print_list() 中定义 char* 的方式进行。

更具体地说，我的问题在 print_list() 中：

printf("\n [%s] \n", ptr -> name);
printf("\n [%s] \n", ptr -> lastn);

所以我的实际代码是（出现 0 个错误和 0 个警告但程序崩溃）：

#include <stdio.h> 
#include <stdlib.h> 
#include <stdbool.h> 

// Creating structure for node 
struct test_struct 
{ 
    int val;         // val is member id number 
    char name; 
    char lastn; 
    int age; 
    struct test_struct *next; 
}; 

// declaring global head and curr pointers 
struct test_struct *head = NULL; 
struct test_struct *curr = NULL; 

// creating a list 
struct test_struct* create_list(int val, char* name, char* lastn, int age) 
{ 
    printf("\n creating list with head node as [%d] [%s] [%s] [%d] \n", val, name, lastn, age); 

    struct test_struct *ptr = malloc(sizeof(struct test_struct)); // creating list 
    if(NULL == ptr) { 
        printf("\n Node creation failed \n"); 
        return NULL; 
    } 

    ptr->val = val; 
    ptr->name = *name; 
    ptr->lastn = *lastn; 
    ptr->age = age; 
    ptr->next = NULL; 

    head = curr = ptr; 

    return ptr; 
}

// add member to list 
struct test_struct* add_to_list(int val, char *name, char *lastn, int age, bool add_to_end) 
{ 
    if(NULL == head) { 
        return (create_list(val, name, lastn, age)); 
    }      

    if(add_to_end) { 
        printf("\n Adding node to end of list with data [%d] [%s] [%s] [%d] \n",  val, name, lastn, age); 
    } else { 
        printf("\n Adding node to beginning of list with data [%d] [%s] [%s] [%d] \n", val, name, lastn, age); 
    } 

    struct test_struct *ptr = malloc(sizeof(struct test_struct)); 

    if(NULL == ptr) { 
        printf("\n Node creation failed \n"); 
        return NULL; 
    } 

    ptr->val = val; 
    ptr->name = *name; 
    ptr->lastn = *lastn; 
    ptr->age = age; 
    ptr->next = NULL; 

    if (add_to_end) { 
        curr-> next = ptr; 
        curr = ptr; 
    } else { 
        ptr -> next = head; 
        head = ptr; 
    } 

    return ptr; 
} 

//printing the list 
void print_list(void) 
{ 
    struct test_struct *ptr = head; 

    printf("\n -----Printing list Start----- \n"); 

    while(ptr != NULL) { 
        printf("\n [%d] \n", ptr -> val); 
        printf("\n [%s] \n", ptr -> name); 
        printf("\n [%s] \n", ptr -> lastn); 
        printf("\n [%d] \n", ptr -> age); 
        ptr = ptr->next; 
    } 

    printf("\n -----Printing list end---- \n"); 

    return; 
}   

// main function 
int main(void) 
{ 
    struct test_struct *ptr = NULL; 

    // for adding member to list 
    add_to_list(123, "william", "shakespeare", 30, true); 
    add_to_list(124, "william", "gibson", 35, true); 
    add_to_list(125, "chuck", "palahniuk", 40, true); 
    add_to_list(126, "mario", "puzio", 50, true); 
    add_to_list(127, "umberto", "eco", 60, true); 
    add_to_list(128, "ezra", "pound", 125, true); 

    print_list(); 

    return 0; 
}

Answer 1

您已将姓名和姓氏声明为单个字符

struct test_struct
{
int val;         // val is member id number
char name;
char lastn;
int age;
struct test_struct *next;
};

您需要将它们声明为固定大小的数组或指向已分配 space 以保存字符串的指针。字符串是以 \0.

结尾的一系列字符

struct test_struct
{
int val;         // val is member id number
char name[MAXLEN];
char lastn[MAXLEN];
int age;
struct test_struct *next;
};

然后将函数的参数复制到结构中的字段

例如

strcpy(ptr->name,name);
strcpy(ptr->lastn,lastn);

Answer 2

printf("\n [%s] \n", ptr -> name);
printf("\n [%s] \n", ptr -> lastn);

%s 期望 char * 而不是 char 因为 name 和 lastn 都是 char 变量。

要存储一个人的名字和姓氏，您应该选择 char array，因为单身 char variable 无法存储它。因此，将它们声明为 char arrays .

例子-

struct test_struct
{
  int val;         // val is member id number
  char name[20];         // or any desired length to store a name 
  char lastn[20];        // similar as for name
  int age;
  struct test_struct *next;
};

然后复制其中的数据使用strncpy-

ptr->name = *name;          // strncpy(ptr->name,name,strlen(name));
ptr->lastn = *lastn;        // strncpy(ptr->lastn,lastn,strlen(lastn));

链表中 char、char* 的区别

Difference between char, char* inside Linked Lists

c

linked-list

char

char-pointer