链表中char，char *之间的区别

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，其他函数将其打印出来。 当我只打印int类型时，我没有任何问题，但是当我尝试也打印char类型时，程序崩溃了。

因此，它必须按照我在打印函数print_list（）中定义char *的方式进行操作。

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（得到0个错误和0个警告，但程序崩溃了）：

#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

您已将名称和名字声明为单个字符

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存储字符串的已分配空间的指针。 字符串是一系列以\\ 0结尾的字符。

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 *不是char因为name和lastn都是char变量。

并且要存储一个人的名字和姓氏，您应该选择char array因为单个char variable不能存储它。 因此，将它们声明为char arrays 。

范例-

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;
};

然后使用strncpy复制数据-

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