为什么我的代码只能打印一个节点值，为什么不能打印其他节点值？

Question

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

struct node{
    int data;
    struct node *next;
};

struct node *root_1 = NULL;
struct node *root_2 = NULL;
struct node *current;

void create(int option){

    struct node *temp,*current;
    int data;

    printf("Enter data value \n");
    scanf("%d",&data);

    if(option == 1){
        if(root_1 == NULL){         
            temp = (struct node*)malloc(sizeof(struct node));
            temp->data = data;
    //      temp->next = Null;
            current = temp; 
            root_1 = temp;
        }else{
            temp = (struct node*)malloc(sizeof(struct node));
            temp->data = data;
            current->next = temp;  
            current = temp; 
            }           
    }else{
        if(root_2 == NULL){         
            temp = (struct node*)malloc(sizeof(struct node));
            temp->data = data;
    //      temp->next = Null;
            current = temp; 
            root_2 = temp;
        }else{
            temp = (struct node*)malloc(sizeof(struct node));
            temp->data = data;
            current->next = temp;  
            current = temp; 
        }           
    }
}

void dispaly(int option){

    printf("List \n");

    struct node *temp;
    if(option == 1){
        temp = root_1;
        while(temp!=NULL){
            printf("%d \n",temp->data);
            temp = temp->next;
        }
    }else{
        temp = root_2;
        while(temp!=NULL){
            printf("%d \n",temp->data);
            temp = temp->next;
        }
    }
}

void function(int option){
        int ch;
        while(1){
            printf("1.create,2.display,3.exit \n");
            printf("Ur choice \n");
            scanf("%d",&ch);

            switch(ch){
                case 1:
                    create(option);
                    break;
                case 2:
                    dispaly(option);
                    break;
                case 3:
                    exit(0);
                default:
                    printf("Wrong option \n");
            }
        }
}

void main(){
    int option;
    while(1){
        printf("1.List-1 \t 2.List-2 \n");
        printf("enter ur option \n");
        scanf("%d",&option);
        switch(option){
            case 1:
                function(option);
                break;
            case 2:
                function(option);
                break;
        }
    }
}

在我的代码中，我试图创建两个链接 lists.in 这个程序我可以创建两个列表，但我无法显示，请任何人帮助 me.two 区分两个列表我正在创建两个根值，它们是 root_1 和 root_2。

Answer 1

// still to do:
// apply the same fix as suggested for the option == 1 code, 
// to the option == 2 code

// Note: in comparison statements, like: 'if(option == 1)',
// the literal should be on the left to enable the compiler to catch
// the following error: 
// 'if( option = 1)'  <-- single '=' rather than '=='
// I.E. use: 'if( 1 == option )'

// IMO: 'georgian' code formatting
//       removes the vertical white space 
//       that is used to greatly enhance the readability of the code
//       suggest always putting '{' and '}' on their own line

struct node *root_1 = NULL;
struct node *root_2 = NULL;
struct node *current;  // <-- remove this line (or rename the variable)
                       //      as 'current' is one of the auto variables
                       //      in the create() function

void create(int option){

    struct node *temp,*current;  
    int data;

    printf("Enter data value \n");
    scanf("%d",&data);          // <-- change to:
                                // if( 1 != scanf( " %d", &data ) )
                                // { // then scanf failed
                                //     perror( "scanf for data failed" );
                                //     // free all malloc'd areas
                                //     // for both linked lists here
                                //     exit( EXIT_FAILURE );
                                //  }
                                //
                                //  // implied else, scanf successful
                                //

    if(option == 1){
        if(root_1 == NULL){         
            temp = (struct node*)malloc(sizeof(struct node));
            temp->data = data;
    //      temp->next = Null; // <-- this line should be: 'temp->next = NULL;'
            current = temp;    // <-- remove this line, it does nothing useful
                               // attach first node to linked list
            root_1 = temp;
        }else{
            temp = (struct node*)malloc(sizeof(struct node));
            temp->data = data;
                                // <-- insert following line to initialize link field
                                temp->next = NULL;
            current = root1;
                                // <-- insert following lines 
                                // step through linked list until end
                                while( NULL != current->next )
                                {
                                    current = current->next;
                                } // end while
                                // attach new node to end of linked list
            current->next = temp; 

            current = temp;     // <-- remove this line, 
                                // <-- it will corrupt the linked list

            }            
    }else{
        if(root_2 == NULL){         
            temp = (struct node*)malloc(sizeof(struct node));
            temp->data = data;
    //      temp->next = Null;
            current = temp; 
            root_2 = temp;
        }else{
            temp = (struct node*)malloc(sizeof(struct node));
            temp->data = data;
            current->next = temp;  
            current = temp; 
        }           
    }
}

为什么我的代码只能打印一个节点值，为什么不能打印其他节点值？

Why my code is able to print only one node value why not other node value?

c

linked-list