Question

我尝试了structur中的结构列表编程和函数中的pointers.passing参数以及访问结构成员，我得到了以下错误请帮忙。

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

    enter code here

 struct node {

    int32_t data;
    struct node *next;
};


void 
     SLL_insert_beg(      struct node *head_lst

                            ){
    int32_t num_lst; 
    struct node *temp_lst,*new_lst;      //intialising  local variable. 
    new_lst = ( struct node* )malloc( sizeof(struct node) );   //Allocating dynamic memory for node new.
    printf(" Enter data: ");
    scanf("%d", &num_lst );
    new_lst->data = num_lst;                   //inserting data into the datafield in node new.

    if(head_lst == NULL)  {                    //Condition to check list is empty in SLL_insert_beg() function.

        new_lst->next = NULL;                  //Making pointer field to point null in node new. 
        head_lst = new_lst;                    //Making new node as head node.

     } else {

        new_lst->next = head_lst;              //pointer field in new node points to the head node.
        head_lst = new_lst;                    //Making new node as head node. 
    }
}

void SLL_display(     struct node *head_lst    


                                                  ) {

        struct node *temp_lst,*new_lst;
        int32_t i_lst = 1 ;                      //intialising  local variable.
    if( head_lst == NULL ) {                 //Condition to check list is empty in  SLL_display() function.

          printf( " List is empty!! " );

    } else {

        temp_lst = head_lst;                 //Making head node as temp node.
        printf( "\nThe linked list is:\n " );
        printf( " Elements: " );
                                             /*Below loop used to print all elements 
                                               in single linked list*/
        while( temp_lst != NULL ) {  

            printf( "%d-> ", temp_lst->data );
            temp_lst = temp_lst->next;       //Making  temp next node as  temp node.
        }
        printf( "NULL\n " );
        printf( " Position: " );

        temp_lst = head_lst;                 //Making head node as temp node.
                                             /*Below loop used to print no.of positions 
                                               in single linked list*/
        while( temp_lst != NULL ) {

            printf( "%d    " , i_lst );
           i_lst = i_lst + 1;                //Incrementing variable 'i_lst' once.
            temp_lst = temp_lst->next;       //Making  temp next node as  temp node.
        }

    }

 }                              
void main(void) {                                                  //sll_program(void) {

    int32_t data;
    int32_t SL_ch; //Intialised a variable  used to choose the option in the below menu.
   struct node *head_lst=NULL;
    struct node *temp_lst,*new_lst;
     while( 1 ) {

                 printf(" \n\nSingly Linked List(SLL)");
                 printf(" \n1.Insert beg     \
                          \n2.diplay  \
                           \n3.exit     "   );
                 printf("\nEnter your choice(1-4):");
                 scanf("%d",&SL_ch);
       switch(SL_ch) {

                     case 1:                  //If choice is 1 calls insert at beginning function.
                             SLL_insert_beg(    &head_lst
                                                              );
                             break;
                     case 2: 
                          SLL_display(           &head_lst



                                                             );                   //If choice is 2 calls the display function.
                                    break;
                      case 3: 
                                 return;
                    default: 
                               printf( "Please give the right choice!!" );              

       }
     }
}

编译后这是我的代码的错误

main.c：在函数'main'中：

main.c：91：63：警告：传递'SLL_insert_beg'的参数1   不兼容的指针类型[默认启用]
  main.c：12：6：注意：预期的'struct node *'但是参数是类型的   'struct node **'         SLL_insert_beg（struct node * head_lst main.c：98：62：警告：从'SLL_display'传递参数1   不兼容的指针类型[默认启用]
  ）; //如果选择2则调用显示功能。

main.c：34：6：注意：预期的'struct node *'但是参数是类型的   'struct node **'   void SLL_display（struct node * head_lst

Answer 1

问题是声明是SLL_insert_beg(struct node *head_lst);，调用是SLL_insert_beg(&head_lst); - 所有这一切都被head_lst声明为指向struct node的指针（struct node *head_lst=NULL ）。这会导致&head_lst成为指向struct node的指针。

换句话说，正如编译器警告：expected 'struct node *' but argument is of type 'struct node **'

更改函数的声明，或更改调用。

结构和指针的概念

1 个答案: